分类: 模板 - CheaSim Blog

模板

Java大数模板

二、Java之输入输出处理
由于ACM竞赛题目的输入数据和输出数据一般有多组(不定),并且格式多种多样,所以,如何处理题目的输入输出是对大家的一项最基本的要求。这也是困扰初学者的一大问题。

  1. 输入:

格式1:Scanner sc = new Scanner (new BufferedInputStream(System.in));

格式2:Scanner sc = new Scanner (System.in);

在读入数据量大的情况下,格式1的速度会快些。

读一个整数: int n = sc.nextInt(); 相当于 scanf(“%d”, &n); 或 cin >> n;

读一个字符串:String s = sc.next(); 相当于 scanf(“%s”, s); 或 cin >> s;

读一个浮点数:double t = sc.nextDouble(); 相当于 scanf(“%lf”, &t); 或 cin >> t;

读一整行: String s = sc.nextLine(); 相当于 gets(s); 或 cin.getline(…);

判断是否有下一个输入可以用sc.hasNext()或sc.hasNextInt()或sc.hasNextDouble()或sc.hasNextLine()

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import java.util.Scanner;
import java.math.*;
Scanner cin = new Scanner (new BufferedInputStream(System.in));
while(cin.haxNextInt){
//操作
}

大数使用例子

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import java.io.*;
import java.math.*;
import java.util.*;

public class Main {
static Scanner in = new Scanner(System.in);
public static void main(String[] args) {
int a = in.nextInt();
BigInteger b = in.nextBigInteger();
BigDecimal c = in.nextBigDecimal();
/*
BigDecimal:
构造方法:
BigDecimal(BigInteger val)
BigDecimal(BigInteger unscaledVal, int scale)
BigDecimal(BigInteger unscaledVal, int scale, MathContext mc)
BigDecimal(BigInteger val, MathContext mc)
BigDecimal(char[] in)
BigDecimal(char[] in, int offset, int len)
BigDecimal(char[] in, int offset, int len, MathContext mc)
BigDecimal(char[] in, MathContext mc)
BigDecimal(double val)
BigDecimal(double val, MathContext mc)
BigDecimal(int val)
BigDecimal(int val, MathContext mc)
BigDecimal(long val)
BigDecimal(long val, MathContext mc)
BigDecimal(String val)
BigDecimal(String val, MathContext mc)
成员方法:
BigDecimal abs()
BigDecimal abs(MathContext mc)
BigDecimal add(BigDecimal augend)
BigDecimal add(BigDecimal augend, MathContext mc)
byte byteValueExact()
int compareTo(BigDecimal val)
BigDecimal divide(BigDecimal divisor)
BigDecimal divide(BigDecimal divisor, int roundingMode)
BigDecimal divide(BigDecimal divisor, int scale, int roundingMode)
BigDecimal divide(BigDecimal divisor, int scale, RoundingMode roundingMode)
BigDecimal divide(BigDecimal divisor, MathContext mc)
BigDecimal divide(BigDecimal divisor, RoundingMode roundingMode)
BigDecimal[] divideAndRemainder(BigDecimal divisor)
BigDecimal[] divideAndRemainder(BigDecimal divisor, MathContext mc)
BigDecimal divideToIntegralValue(BigDecimal divisor)
BigDecimal divideToIntegralValue(BigDecimal divisor, MathContext mc)
double doubleValue()
boolean equals(Object x)
float floatValue()
int hashCode()
int intValue()
int intValueExact()
long longValue()
long longValueExact()
BigDecimal max(BigDecimal val)
BigDecimal min(BigDecimal val)
BigDecimal movePointLeft(int n)
BigDecimal movePointRight(int n)
BigDecimal multiply(BigDecimal multiplicand)
BigDecimal multiply(BigDecimal multiplicand, MathContext mc)
BigDecimal negate()
BigDecimal negate(MathContext mc)
BigDecimal plus()
BigDecimal plus(MathContext mc)
BigDecimal pow(int n)
BigDecimal pow(int n, MathContext mc)
int precision()
BigDecimal remainder(BigDecimal divisor)
BigDecimal remainder(BigDecimal divisor, MathContext mc)
BigDecimal round(MathContext mc)
int scale()
BigDecimal scaleByPowerOfTen(int n)
BigDecimal setScale(int newScale)
Returns a BigDecimal whose scale is the specified value, and whose value is numerically equal to this BigDecimal's.
BigDecimal setScale(int newScale, int roundingMode)
BigDecimal setScale(int newScale, RoundingMode roundingMode)
Returns a BigDecimal whose scale is the specified value, and whose unscaled value is determined by multiplying or dividing this BigDecimal's unscaled value by the appropriate power of ten to maintain its overall value.
short shortValueExact()
int signum()
Returns the signum function of this BigDecimal. (1,0,-1)
BigDecimal stripTrailingZeros()
Returns a BigDecimal which is numerically equal to this one but with any trailing zeros removed from the representation.
BigDecimal subtract(BigDecimal subtrahend)
BigDecimal subtract(BigDecimal subtrahend, MathContext mc)
BigInteger toBigInteger()
BigInteger toBigIntegerExact()
String toEngineeringString()
String toPlainString()
String toString()
BigDecimal ulp()
Returns the size of an ulp, a unit in the last place, of this BigDecimal.
BigInteger unscaledValue()
static BigDecimal valueOf(double val)
static BigDecimal valueOf(long val)
static BigDecimal valueOf(long unscaledVal, int scale)
*/
BigDecimal test = new BigDecimal("1.234567");
test = test.setScale(3, RoundingMode.HALF_UP);
System.out.println(test);
test = test.setScale(7, RoundingMode.HALF_EVEN);
System.out.println(test);
test = test.divide(new BigDecimal("3"), MathContext.UNLIMITED); // 默认也是UNLIMITED精度,无限小数会报错
System.out.println(test);

/*
BigInteger:
构造方法:
BigInteger(byte[] val)
BigInteger(int signum, byte[] magnitude)
BigInteger(int bitLength, int certainty, Random rnd)
Constructs a randomly generated positive BigInteger that is probably prime, with the specified bitLength.
BigInteger(int numBits, Random rnd)
Constructs a randomly generated BigInteger, uniformly distributed over the range 0 to (2numBits - 1), inclusive.
BigInteger(String val)
BigInteger(String val, int radix)
成员方法:
BigInteger abs()
BigInteger add(BigInteger val)
BigInteger and(BigInteger val)
BigInteger andNot(BigInteger val)
Returns a BigInteger whose value is (this & ~val).
int bitCount()
Returns the number of bits in the two's complement representation of this BigInteger that differ from its sign bit.
int bitLength()
Returns the number of bits in the minimal two's-complement representation of this BigInteger, excluding a sign bit.
BigInteger clearBit(int n)
Returns a BigInteger whose value is equivalent to this BigInteger with the designated bit cleared.
int compareTo(BigInteger val)
BigInteger divide(BigInteger val)
BigInteger[] divideAndRemainder(BigInteger val)
double doubleValue()
boolean equals(Object x)
BigInteger flipBit(int n)
Returns a BigInteger whose value is equivalent to this BigInteger with the designated bit flipped.
float floatValue()
BigInteger gcd(BigInteger val)
Returns a BigInteger whose value is the greatest common divisor of abs(this) and abs(val).
int getLowestSetBit()
Returns the index of the rightmost (lowest-order) one bit in this BigInteger (the number of zero bits to the right of the rightmost one bit).
int hashCode()
int intValue()
boolean isProbablePrime(int certainty)
Returns true if this BigInteger is probably prime, false if it's definitely composite.
long longValue()
BigInteger max(BigInteger val)
BigInteger min(BigInteger val)
BigInteger mod(BigInteger m)
BigInteger modInverse(BigInteger m)
Returns a BigInteger whose value is (this^-1 mod m).
BigInteger modPow(BigInteger exponent, BigInteger m)
BigInteger multiply(BigInteger val)
BigInteger negate()
BigInteger nextProbablePrime()
Returns the first integer greater than this BigInteger that is probably prime.
BigInteger not()
BigInteger or(BigInteger val)
BigInteger pow(int exponent)
static BigInteger probablePrime(int bitLength, Random rnd)
Returns a positive BigInteger that is probably prime, with the specified bitLength.
BigInteger remainder(BigInteger val)
Returns a BigInteger whose value is (this % val).
BigInteger setBit(int n)
Returns a BigInteger whose value is equivalent to this BigInteger with the designated bit set.
BigInteger shiftLeft(int n)
Returns a BigInteger whose value is (this << n).
BigInteger shiftRight(int n)
Returns a BigInteger whose value is (this >> n).
int signum()
BigInteger subtract(BigInteger val)
boolean testBit(int n)
Returns true if and only if the designated bit is set.
byte[] toByteArray()
String toString()
String toString(int radix)
static BigInteger valueOf(long val)
BigInteger xor(BigInteger val)
*/

MyPair[] pairs = new MyPair[1000];
Arrays.sort(pairs);
Arrays.binarySearch(pairs, 1, 4, new MyPair());
/*
二分查找
如果元素在数组中,则值为0~n-1,否则值为-1~-(n+1),表示第一个比它大的值的位置,下标从1开始
*/
List<MyPair> pairList = new ArrayList<>();
pairList.add(new MyPair());
//pairList.sort();
pairList.sort(new Cmp());
Collections.shuffle(pairList);
Collections.swap(pairList, 1, 3);
Collections.sort(pairList);
}
}

class MyPair implements Comparable {
int x, y;

@Override
public int compareTo(Object o) {
MyPair b = (MyPair)o;
if(x!=b.x) return x<b.x?-1:1;
else if(y!=b.y) return y<b.y?-1:1;
return 0;
}
}

class Cmp implements Comparator<MyPair> {
@Override
public int compare(MyPair o1, MyPair o2) {
if(o1.x!=o2.x) return o1.x<o2.x?-1:1;
else if(o1.y!=o2.y) return o1.y<o2.y?-1:1;
return 0;
}
}

关于树的

树链剖分

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#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for(int i=(a);i<(n);i++)
#define per(i,a,n) for(int i=(n-1);i>=(a);i--)
#define fi first
#define se second
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
void swap(int &a,int &b){
int temp = a;
a = b;
b = temp;
}
typedef pair <int,int> pII;
typedef long long ll;
const int INF = 0x3f3f3f3f;
//head
const int maxn = 5e4+10;
int size[maxn],dep[maxn],top[maxn],fa[maxn],id[maxn],son[maxn];
int head[maxn],w[maxn];
int n,cnt,totw,m;
struct edge{
int next,to;
}G[maxn<<1];
int find(int x){ return x==fa[x]?x:find(fa[x]);}
void addedge(int u,int v){
G[cnt].next = head[u];
G[cnt].to = v;
head[u] = cnt++;
}
void dfs1(int u,int f){
size[u] = 1;
for(int i=head[u];~i;i=G[i].next){
int v = G[i].to;
if(v==f) continue;
dep[v] = dep[u]+1;
fa[v] = u;
dfs1(v,u);
if(size[v] > size[son[u]]) son[u] = v;
size[u]+=size[v];
}
}
void dfs2(int u,int topu){
top[u] = topu;
id[u] = ++totw;
if(son[u]) dfs2(son[u],top[u]);
for(int i=head[u];~i;i=G[i].next){
int v = G[i].to;
if(v==fa[u] || v==son[u]) continue;
dfs2(v,v);
}
}
int data[maxn<<2],lazy[maxn<<2];
void pushdown(int rt){
if(lazy[rt]!=0){
data[rt<<1] += lazy[rt];
data[rt<<1|1] += lazy[rt];
lazy[rt<<1] += lazy[rt];
lazy[rt<<1|1] += lazy[rt];
lazy[rt] = 0;
}
}
void build(int l,int r,int rt){
lazy[rt] = 0;
if(l==r){
data[rt] = w[l];
return;
}
int mid = l+r>>1;
build(lson); build(rson);
}
int query(int pos,int l,int r,int rt){
if(l==r && l==pos) return data[rt];
int mid = l+r>>1;
pushdown(rt);
if(pos<=mid) return query(pos,lson);
else return query(pos,rson);
}
void update(int val,int L,int R,int l,int r,int rt){
if(L<=l && r<=R){
lazy[rt] += val;
data[rt] += val;
return;
}
pushdown(rt);
int mid = l+r>>1;
if(L<=mid) update(val,L,R,lson);
if(mid<R) update(val,L,R,rson);
}
void change(int u,int v,int val){
int t1=top[u],t2=top[v];
while(t1!=t2){
if(dep[t1]<dep[t2]){
swap(t1,t2);
swap(u,v);
}
update(val,id[t1],id[u],1,totw,1);
u = fa[t1];
t1 = top[u];
}
if(dep[u]>dep[v]) swap(u,v);
update(val,id[u],id[v],1,totw,1);
}
int val[maxn];
int p;
char op[20];
void init(){
cnt = totw = 0;
memset(head,-1,sizeof(head));
dep[1] = fa[1] = size[0] = 0;
memset(son,0,sizeof(son));
}
int main(){
#ifdef LOCAL
freopen("4.in","r",stdin);
#endif
while(~scanf("%d%d%d",&n,&m,&p)){
init();
rep(i,1,n+1) scanf("%d",val+i);
rep(i,0,m){
int u,v;scanf("%d%d",&u,&v);
addedge(u,v); addedge(v,u);
}
dfs1(1,0);
dfs2(1,1);
rep(i,1,n+1) w[id[i]] = val[i];
build(1,totw,1);
while(p--){
scanf("%s",op);
if(op[0]=='Q'){
int x;scanf("%d",&x);
printf("%d\n",query(id[x],1,totw,1));
}else{
int u,v,val;scanf("%d%d%d",&u,&v,&val);
if(op[0]=='D') val = -val;
change(u,v,val);
}
}
}

return 0;
}

线段树

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#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for(int i=(a);i<(n);i++)
#define per(i,a,n) for(int i=(n-1);i>=(a);i--)
#define fi first
#define se second
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
typedef pair <int,int> pII;
typedef long long ll;
const int INF = 0x3f3f3f3f;
//head ----------------------------------------
const int maxn = 1e6+20;
int T,n,m;
struct sgt{
int maxlen[maxn<<2];
int ll[maxn<<2],rr[maxn<<2],mm[maxn<<2];
int lazy[maxn<<2];
void build(int l,int r,int rt){
lazy[rt] = -1;
if(l==r){
ll[rt] = rr[rt] = 0;
mm[rt] = 1;
return;
}
int mid = l+r>>1;
build(lson); build(rson);
}
void init(){
lazy[1] = 1;
}
void pushup(int l,int r,int rt){
if(lazy[rt]>=0){
maxlen[rt] = ll[rt] = rr[rt] = mm[rt] = (lazy[rt]?(r-l+1):0);
}else{
int lc = rt<<1,rc = rt<<1|1,mid = l+r>>1;
ll[rt] = ll[lc];
if(ll[lc]==(mid-l+1)) ll[rt] += ll[rc];
mm[rt] = rr[lc] + ll[rc];
rr[rt] = rr[rc];
if(rr[rc] == r-mid) rr[rt] += rr[lc];
maxlen[rt] = max(maxlen[lc],max(maxlen[rc],mm[rt]));
}
}
void pushdown(int rt){
if(lazy[rt]>=0){
int lc=rt<<1,rc=rt<<1|1;
lazy[lc] = lazy[rc] = lazy[rt];
lazy[rt] = -1;
}
}
int query(int val,int l,int r,int rt){
pushup(l,r,rt);
//cout<<l<<' '<<r<<' '<<" ";
int lc = rt<<1,rc = rt<<1|1,mid = l+r>>1;
if(val>maxlen[rt]) return 0;
if(ll[rt] >= val) return l;
pushdown(rt);
pushup(rson);
int temp = query(val,lson);
if(temp) return temp;
if(mm[rt] >= val) return mid-rr[lc]+1;
return query(val,rson);
}
void update(int val,int L,int R,int l,int r,int rt){
if(L<=l && r<=R){
lazy[rt] = val;
pushup(l,r,rt);
return;
}
pushdown(rt);
int mid = l+r>>1;
if(L<=mid) update(val,L,R,lson);
else pushup(lson);
if(mid<R) update(val,L,R,rson);
else pushup(rson);
pushup(l,r,rt);
}
}all,ns;
char op[20];
int main(){
#ifdef LOCAL
freopen("1.in","r",stdin);
#endif
scanf("%d",&T);
rep(test_case,1,T+1){
printf("Case %d:\n",test_case);
scanf("%d%d",&n,&m);
//all.build(1,n,1); ns.build(1,n,1);
all.init(); ns.init();
while(m--){
scanf("%s",op);
if(op[0]=='D'){
int x;scanf("%d",&x);
int ans = all.query(x,1,n,1);
if(ans){
all.update(0,ans,ans+x-1,1,n,1);
printf("%d,let's fly\n",ans);
}else{
puts("fly with yourself");
}
}else if(op[0]=='N'){
int x;scanf("%d",&x);
int ans = all.query(x,1,n,1);
if(!ans) ans = ns.query(x,1,n,1);
if(ans){
all.update(0,ans,ans+x-1,1,n,1);
ns.update(0,ans,ans+x-1,1,n,1);
printf("%d,don't put my gezi\n",ans);
}else{
puts("wait for me");
}
}else{
int x,y;scanf("%d%d",&x,&y);
all.update(1,x,y,1,n,1);
ns.update(1,x,y,1,n,1);
puts("I am the hope of chinese chengxuyuan!!");
}
}
}

return 0;
}

LCA 最近公共父节点 用来求两点之间距离的

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#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for(int i=(a);i<(n);i++)
#define per(i,a,n) for(int i=(n-1);i>=(a);i--)
#define fi first
#define se second
typedef pair <int,int> pII;
typedef long long ll;
const int INF = 0x3f3f3f3f;
//head
const int maxn = 1e5 + 10;
int cnt,n,m;
int head[maxn];
int fa[maxn];
struct node{
int next,to;
}G[maxn<<1];
void addedge(int u,int v){
G[cnt].to = v;
G[cnt].next = head[u];
head[u] = cnt++;
}
struct ST{
int tot;
int first[maxn<<1],R[maxn<<1],order[maxn<<1],dp[maxn<<1][20],dis[maxn];
void init(int root){
tot = 0;
dfs(root,0,-1);
ST_init(tot);
}
void ST_init(int n){
rep(i,1,n+1) dp[i][0] = i;
for(int j=1;(1<<j)<=n;j++){
for(int i=1;i+(1<<j)-1<=n;i++){
int a = dp[i][j-1],b = dp[i+(1<<(j-1))][j-1];
dp[i][j] = R[a]<R[b]?a:b;
}
}
}
int LCA(int u,int v){
int x = first[u],y = first[v];
if(x>y) swap(x,y);
int res = RMQ(x,y);
return order[res];
}
int RMQ(int l,int r){
int k = 0;
while((1<<(k+1))<=r-l+1) k++;
int a = dp[l][k],b = dp[r-(1<<k)+1][k];
return R[a]<R[b]?a:b;
}
void dfs(int u,int deep,int f){
R[++tot] = deep;
dis[u] = deep;
order[tot] = u;
first[u] = tot;
for(int i=head[u];~i;i=G[i].next){
int v = G[i].to;
if(v==f) continue;
dfs(v,deep+1,u);
order[++tot] = u;
R[tot] = deep;
}
}
}st;
int find(int x){
return x==fa[x]?x:fa[x] = find(fa[x]);
}
int ind[maxn];
char s[60];
void init(){
cnt = 0;
memset(head,-1,sizeof(head));
memset(ind,0,sizeof(ind));
}
int main(){
#ifdef LOCAL
freopen("2.in","r",stdin);
#endif
int T;scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&m);
init();
map<string,int> mp;
int tot = 0;
rep(i,0,n-1){
int u,v;
scanf("%s",s);
if(mp[s]==0) mp[s] = ++tot,u = tot;
else u = mp[s];
scanf("%s",s);
if(mp[s]==0) mp[s] = ++tot,v = tot;
else v = mp[s];
addedge(v,u); ind[u]++;
}
int root = 0;
rep(i,1,n+1) if(ind[i]==0){
root = i;
break;
}
st.init(root);
rep(i,0,m){
int u,v,ans;
scanf("%s",s); u = mp[s];
scanf("%s",s); v = mp[s];
int rt = st.LCA(u,v);
int x = st.dis[u] - st.dis[rt];
int y = st.dis[v] - st.dis[rt];
if(rt==u) ans = 1;
else if(rt == v) ans = x;
else ans = x+1;
if(u==v) ans = 0;
printf("%d\n",ans);
}
}
return 0;
}

计算几何模板

J .Distance to work

题意

懒得看了

题解

计算几何。

扒一扒dls的板子

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typedef double db;
const db EPS = 1e-9;

inline int sign(db a) { return a < -EPS ? -1 : a > EPS; }

inline int cmp(db a, db b){ return sign(a-b); }

struct P {
db x, y;
P() {}
P(db _x, db _y) : x(_x), y(_y) {}
P operator+(P p) { return {x + p.x, y + p.y}; }
P operator-(P p) { return {x - p.x, y - p.y}; }
P operator*(db d) { return {x * d, y * d}; }
P operator/(db d) { return {x / d, y / d}; }

bool operator<(P p) const {
int c = cmp(x, p.x);
if (c) return c == -1;
return cmp(y, p.y) == -1;
}

bool operator==(P o) const{
return cmp(x,o.x) == 0 && cmp(y,o.y) == 0;
}

db dot(P p) { return x * p.x + y * p.y; }
db det(P p) { return x * p.y - y * p.x; }

db distTo(P p) { return (*this-p).abs(); }
db alpha() { return atan2(y, x); }
void read() { cin>>x>>y; }
void write() {cout<<"("<<x<<","<<y<<")"<<endl;}
db abs() { return sqrt(abs2());}
db abs2() { return x * x + y * y; }
P rot90() { return P(-y,x);}
P unit() { return *this/abs(); }
int quad() const { return sign(y) == 1 || (sign(y) == 0 && sign(x) >= 0); }
P rot(db an){ return {x*cos(an)-y*sin(an),x*sin(an) + y*cos(an)}; }
};

struct L{ //ps[0] -> ps[1]
P ps[2];
L() {}
L(P p1,P p2) { ps[0]=p1; ps[1]=p2; }
P& operator[](int i) { return ps[i]; }
P dir() { return ps[1] - ps[0]; }
bool include(P p) { return sign((ps[1] - ps[0]).det(p - ps[0])) > 0; }
L push(){ // push eps outward
const double eps = 1e-6;
P delta = (ps[1] - ps[0]).rot90().unit() * eps;
return L(ps[0] - delta, ps[1] - delta);
}
};
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#define cross(p1,p2,p3) ((p2.x-p1.x)*(p3.y-p1.y)-(p3.x-p1.x)*(p2.y-p1.y))
#define crossOp(p1,p2,p3) sign(cross(p1,p2,p3))

bool chkLL(P p1, P p2, P q1, P q2) {
db a1 = cross(q1, q2, p1), a2 = -cross(q1, q2, p2);
return sign(a1+a2) != 0;
}

P isLL(P p1, P p2, P q1, P q2) {
db a1 = cross(q1, q2, p1), a2 = -cross(q1, q2, p2);
return (p1 * a2 + p2 * a1) / (a1 + a2);
}

P isLL(L l1,L l2){ return isLL(l1[0],l1[1],l2[0],l2[1]); }

bool intersect(db l1,db r1,db l2,db r2){
if(l1>r1) swap(l1,r1); if(l2>r2) swap(l2,r2);
return !( cmp(r1,l2) == -1 || cmp(r2,l1) == -1 );
}

bool isSS(P p1, P p2, P q1, P q2){
return intersect(p1.x,p2.x,q1.x,q2.x) && intersect(p1.y,p2.y,q1.y,q2.y) &&
crossOp(p1,p2,q1) * crossOp(p1,p2,q2) <= 0 && crossOp(q1,q2,p1)
* crossOp(q1,q2,p2) <= 0;
}

bool isSS_strict(P p1, P p2, P q1, P q2){
return crossOp(p1,p2,q1) * crossOp(p1,p2,q2) < 0 && crossOp(q1,q2,p1)
* crossOp(q1,q2,p2) < 0;
}

bool isMiddle(db a, db m, db b) {
return sign(a - m) == 0 || sign(b - m) == 0 || (a < m != b < m);
}

bool isMiddle(P a, P m, P b) {
return isMiddle(a.x, m.x, b.x) && isMiddle(a.y, m.y, b.y);
}

bool onSeg(P p1, P p2, P q){
return crossOp(p1,p2,q) == 0 && isMiddle(p1, q, p2);
}

bool onSeg_strict(P p1, P p2, P q){
return crossOp(p1,p2,q) == 0 && sign((q-p1).dot(p1-p2)) * sign((q-p2).dot(p1-p2)) < 0;
}

P proj(P p1, P p2, P q) {
P dir = p2 - p1;
return p1 + dir * (dir.dot(q - p1) / dir.abs2());
}

P reflect(P p1, P p2, P q){
return proj(p1,p2,q) * 2 - q;
}

db nearest(P p1,P p2,P q){
P h = proj(p1,p2,q);
if(isMiddle(p1,h,p2))
return q.distTo(h);
return min(p1.distTo(q),p2.distTo(q));
}

db disSS(P p1, P p2, P q1, P q2){
if(isSS(p1,p2,q1,q2)) return 0;
return min(min(nearest(p1,p2,q1),nearest(p1,p2,q2)), min(nearest(q1,q2,p1),nearest(q1,q2,p2)));
}

db rad(P p1,P p2){
return atan2l(p1.det(p2),p1.dot(p2));
}

db incircle(P p1, P p2, P p3){
db A = p1.distTo(p2);
db B = p2.distTo(p3);
db C = p3.distTo(p1);
return sqrtl(A*B*C/(A+B+C));
}

//polygon
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db area(vector<P> ps){
db ret = 0; rep(i,0,ps.size()) ret += ps[i].det(ps[(i+1)%ps.size()]);
return ret/2;
}

int contain(vector<P> ps, P p){ //2:inside,1:on_seg,0:outside
int n = ps.size(), ret = 0;
rep(i,0,n){
P u=ps[i],v=ps[(i+1)%n];
if(onSeg(u,v,p)) return 1;
if(cmp(u.y,v.y)<=0) swap(u,v);
if(cmp(p.y,u.y) >0 || cmp(p.y,v.y) <= 0) continue;
ret ^= crossOp(p,u,v) > 0;
}
return ret*2;
}

vector<P> convexHull(vector<P> ps) {
int n = ps.size(); if(n <= 1) return ps;
sort(ps.begin(), ps.end());
vector<P> qs(n * 2); int k = 0;
for (int i = 0; i < n; qs[k++] = ps[i++])
while (k > 1 && crossOp(qs[k - 2], qs[k - 1], ps[i]) <= 0) --k;
for (int i = n - 2, t = k; i >= 0; qs[k++] = ps[i--])
while (k > t && crossOp(qs[k - 2], qs[k - 1], ps[i]) <= 0) --k;
qs.resize(k - 1);
return qs;
}

vector<P> convexHullNonStrict(vector<P> ps) {
//caution: need to unique the Ps first
int n = ps.size(); if(n <= 1) return ps;
sort(ps.begin(), ps.end());
vector<P> qs(n * 2); int k = 0;
for (int i = 0; i < n; qs[k++] = ps[i++])
while (k > 1 && crossOp(qs[k - 2], qs[k - 1], ps[i]) < 0) --k;
for (int i = n - 2, t = k; i >= 0; qs[k++] = ps[i--])
while (k > t && crossOp(qs[k - 2], qs[k - 1], ps[i]) < 0) --k;
qs.resize(k - 1);
return qs;
}

db convexDiameter(vector<P> ps){
int n = ps.size(); if(n <= 1) return 0;
int is = 0, js = 0; rep(k,1,n) is = ps[k]<ps[is]?k:is, js = ps[js] < ps[k]?k:js;
int i = is, j = js;
db ret = ps[i].distTo(ps[j]);
do{
if((ps[(i+1)%n]-ps[i]).det(ps[(j+1)%n]-ps[j]) >= 0)
(++j)%=n;
else
(++i)%=n;
ret = max(ret,ps[i].distTo(ps[j]));
}while(i!=is || j!=js);
return ret;
}

vector<P> convexCut(const vector<P>&ps, P q1, P q2) {
vector<P> qs;
int n = ps.size();
rep(i,0,n){
P p1 = ps[i], p2 = ps[(i+1)%n];
int d1 = crossOp(q1,q2,p1), d2 = crossOp(q1,q2,p2);
if(d1 >= 0) qs.pb(p1);
if(d1 * d2 < 0) qs.pb(isLL(p1,p2,q1,q2));
}
return qs;
}

//min_dist
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db min_dist(vector<P>&ps,int l,int r){
if(r-l<=5){
db ret = 1e100;
rep(i,l,r) rep(j,l,i) ret = min(ret,ps[i].distTo(ps[j]));
return ret;
}
int m = (l+r)>>1;
db ret = min(min_dist(ps,l,m),min_dist(ps,m,r));
vector<P> qs; rep(i,l,r) if(abs(ps[i].x-ps[m].x)<= ret) qs.pb(ps[i]);
sort(qs.begin(), qs.end(),[](P a,P b) -> bool {return a.y<b.y; });
rep(i,1,qs.size()) for(int j=i-1;j>=0&&qs[j].y>=qs[i].y-ret;--j)
ret = min(ret,qs[i].distTo(qs[j]));
return ret;
}

int type(P o1,db r1,P o2,db r2){
db d = o1.distTo(o2);
if(cmp(d,r1+r2) == 1) return 4;
if(cmp(d,r1+r2) == 0) return 3;
if(cmp(d,abs(r1-r2)) == 1) return 2;
if(cmp(d,abs(r1-r2)) == 0) return 1;
return 0;
}

vector<P> isCL(P o,db r,P p1,P p2){
db x = (p1-o).dot(p2-p1), y = (p2-p1).abs2(), d = x * x - y * ((p1-o).abs2() - r*r);
if(sign(d) < 0) return {};
d = max(d,0.0); P m = p1 - (p2-p1)*(x/y), dr = (p2-p1)*(sqrt(d)/y);
return {m-dr,m+dr}; //along dir: p1->p2
}

vector<P> isCC(P o1, db r1, P o2, db r2) { //need to check whether two circles are the same
db d = o1.distTo(o2);
if (cmp(d, r1 + r2) == 1) return {};
if (cmp(d,abs(r1-r2))==-1) return {};
d = min(d, r1 + r2);
db y = (r1 * r1 + d * d - r2 * r2) / (2 * d), x = sqrt(r1 * r1 - y * y);
P dr = (o2 - o1).unit();
P q1 = o1 + dr * y, q2 = dr.rot90() * x;
return {q1-q2,q1+q2};//along circle 1
}

vector<P> tanCP(P o, db r, P p) {
db x = (p - o).abs2(), d = x - r * r;
if (sign(d) <= 0) return {}; // on circle => no tangent
P q1 = o + (p - o) * (r * r / x);
P q2 = (p - o).rot90() * (r * sqrt(d) / x);
return {q1-q2,q1+q2}; //counter clock-wise
}


vector<L> extanCC(P o1, db r1, P o2, db r2) {
vector<L> ret;
if (cmp(r1, r2) == 0) {
P dr = (o2 - o1).unit().rot90() * r1;
ret.pb(L(o1 + dr, o2 + dr)), ret.pb(L(o1 - dr, o2 - dr));
} else {
P p = (o2 * r1 - o1 * r2) / (r1 - r2);
vector<P> ps = tanCP(o1, r1, p), qs = tanCP(o2, r2, p);
rep(i,0,min(ps.size(),qs.size())) ret.pb(L(ps[i], qs[i])); //c1 counter-clock wise
}
return ret;
}

vector<L> intanCC(P o1, db r1, P o2, db r2) {
vector<L> ret;
P p = (o1 * r2 + o2 * r1) / (r1 + r2);
vector<P> ps = tanCP(o1,r1,p), qs = tanCP(o2,r2,p);
rep(i,0,min(ps.size(),qs.size())) ret.pb(L(ps[i], qs[i])); //c1 counter-clock wise
return ret;
}

db areaCT(db r, P p1, P p2){
vector<P> is = isCL(P(0,0),r,p1,p2);
if(is.empty()) return r*r*rad(p1,p2)/2;
bool b1 = cmp(p1.abs2(),r*r) == 1, b2 = cmp(p2.abs2(), r*r) == 1;
if(b1 && b2){
if(sign((p1-is[0]).dot(p2-is[0])) <= 0 &&
sign((p1-is[0]).dot(p2-is[0])) <= 0)
return r*r*(rad(p1,is[0]) + rad(is[1],p2))/2 + is[0].det(is[1])/2;
else return r*r*rad(p1,p2)/2;
}
if(b1) return (r*r*rad(p1,is[0]) + is[0].det(p2))/2;
if(b2) return (p1.det(is[1]) + r*r*rad(is[1],p2))/2;
return p1.det(p2)/2;
}

bool parallel(L l0, L l1) { return sign( l0.dir().det( l1.dir() ) ) == 0; }

bool sameDir(L l0, L l1) { return parallel(l0, l1) && sign(l0.dir().dot(l1.dir()) ) == 1; }

bool cmp (P a, P b) {
if (a.quad() != b.quad()) {
return a.quad() < b.quad();
} else {
return sign( a.det(b) ) > 0;
}
}

bool operator < (L l0, L l1) {
if (sameDir(l0, l1)) {
return l1.include(l0[0]);
} else {
return cmp( l0.dir(), l1.dir() );
}
}

bool check(L u, L v, L w) {
return w.include(isLL(u,v));
}

vector<P> halfPlaneIS(vector<L> &l) {
sort(l.begin(), l.end());
deque<L> q;
for (int i = 0; i < (int)l.size(); ++i) {
if (i && sameDir(l[i], l[i - 1])) continue;
while (q.size() > 1 && !check(q[q.size() - 2], q[q.size() - 1], l[i])) q.pop_back();
while (q.size() > 1 && !check(q[1], q[0], l[i])) q.pop_front();
q.push_back(l[i]);
}
while (q.size() > 2 && !check(q[q.size() - 2], q[q.size() - 1], q[0])) q.pop_back();
while (q.size() > 2 && !check(q[1], q[0], q[q.size() - 1])) q.pop_front();
vector<P> ret;
for (int i = 0; i < (int)q.size(); ++i) ret.push_back(isLL(q[i], q[(i + 1) % q.size()]));
return ret;
}

P inCenter(P A, P B, P C) {
double a = (B - C).abs(), b = (C - A).abs(), c = (A - B).abs();
return (A * a + B * b + C * c) / (a + b + c);
}

P circumCenter(P a, P b, P c) {
P bb = b - a, cc = c - a;
double db = bb.abs2(), dc = cc.abs2(), d = 2 * bb.det(cc);
return a - P(bb.y * dc - cc.y * db, cc.x * db - bb.x * dc) / d;
}

P othroCenter(P a, P b, P c) {
P ba = b - a, ca = c - a, bc = b - c;
double Y = ba.y * ca.y * bc.y,
A = ca.x * ba.y - ba.x * ca.y,
x0 = (Y + ca.x * ba.y * b.x - ba.x * ca.y * c.x) / A,
y0 = -ba.x * (x0 - c.x) / ba.y + ca.y;
return {x0, y0};
}

AC代码(别人家的模板。)

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#include <bits/stdc++.h>
using namespace std; // 计算几何模板
#define REP(i, x, n) for (int i = x; i < n; ++i)
const double eps = 1e-8;
const double inf = 1e20;
const double pi = acos(-1.0);
const int maxp = 1010;
// Compares a double to zero
int sgn(double x) {
if (fabs(x) < eps) return 0;
if (x < 0)
return -1;
else
return 1;
}
// square of a double
inline double sqr(double x) {
return x * x;
}
/*
15 * Point
16 * Point() - Empty constructor
17 * Point(double _x,double _y) - constructor
18 * input() - double input
19 * output() - %.2f output
20 * operator == - compares x and y
21 * operator < - compares first by x, then by y
22 * operator - - return new Point after subtracting
curresponging x and y
23 * operator ^ - cross product of 2d points
24 * operator * - dot product
25 * len() - gives length from origin
26 * len2() - gives square of length from origin
27 * distance(Point p) - gives distance from p
28 * operator + Point b - returns new Point after adding
curresponging x and y
29 * operator * double k - returns new Point after multiplieing x and
y by k
30 * operator / double k - returns new Point after divideing x and y
by k
31 * rad(Point a,Point b) - returns the angle of Point a and Point b
from this Point
32 * trunc(double r) - return Point that if truncated the
distance from center to r
33 * rotleft() - returns 90 degree ccw rotated point
34 * rotright() - returns 90 degree cw rotated point
35 * rotate(Point p,double angle) - returns Point after rotateing the
Point centering at p by angle radian ccw
36 */
struct Point {
double x, y;
Point() {}
Point(double _x, double _y) {
x = _x;
y = _y;
}
void input() {
scanf("%lf%lf", &x, &y);
}
void output() {
printf("%.2f␣%.2f\n", x, y);
}
bool operator==(Point b) const {
return sgn(x - b.x) == 0 && sgn(y - b.y) == 0;
}
bool operator<(Point b) const {
return sgn(x - b.x) == 0 ? sgn(y - b.y) < 0 : x < b.x;
}
Point operator-(const Point &b) const {
return Point(x - b.x, y - b.y);
}
//叉积
double operator^(const Point &b) const {
return x * b.y - y * b.x;
}
//点积
double operator*(const Point &b) const {
return x * b.x + y * b.y;
}
//返回长度
double len() {
return hypot(x, y); //库函数
}
//返回长度的平方
double len2() {
return x * x + y * y;
}
//返回两点的距离
double distance(Point p) {
return hypot(x - p.x, y - p.y);
}
Point operator+(const Point &b) const {
return Point(x + b.x, y + b.y);
}
Point operator*(const double &k) const {
return Point(x * k, y * k);
}
Point operator/(const double &k) const {
return Point(x / k, y / k);
}
//计算 pa 和 pb 的夹角
//就是求这个点看 a,b 所成的夹角
//测试 LightOJ1203
double rad(Point a, Point b) {
Point p = *this;
return fabs(atan2(fabs((a - p) ^ (b - p)), (a - p) * (b - p)));
}
//化为长度为 r 的向量
Point trunc(double r) {
double l = len();
if (!sgn(l)) return *this;
r /= l;
return Point(x * r, y * r);
}
//逆时针旋转 90 度
Point rotleft() {
return Point(-y, x);
}
//顺时针旋转 90 度
Point rotright() {
return Point(y, -x);
}
//绕着 p 点逆时针旋转 angle
Point rotate(Point p, double angle) {
Point v = (*this) - p;
double c = cos(angle), s = sin(angle);
return Point(p.x + v.x * c - v.y * s, p.y + v.x * s + v.y * c);
}
};
/*
* Stores two points
* Line() - Empty constructor
* Line(Point _s,Point _e) - Line through _s and _e
* operator == - checks if two points are same
* Line(Point p,double angle) - one end p , another end at angle degree
* Line(double a,double b,double c) - Line of equation ax + by + c = 0
* input() - inputs s and e
* adjust() - orders in such a way that s < e
* length() - distance of se
* angle() - return 0 <= angle < pi
* relation(Point p) - 3 if point is on line
* 1 if point on the left of line
* 2 if point on the right of line
* pointonseg(double p) - return true if point on segment
* parallel(Line v) - return true if they are parallel
* segcrossseg(Line v) - returns 0 if does not intersect
* returns 1 if non - standard intersection
* returns 2 if intersects
* linecrossseg(Line v) - line and seg
* linecrossline(Line v) - 0 if parallel
* 1 if coincides
* 2 if intersects
* crosspoint(Line v) - returns intersection point
* dispointtoline(Point p) - distance from point p to the line
* dispointtoseg(Point p) - distance from p to the segment
* dissegtoseg(Line v) - distance of two segment
* lineprog(Point p) - returns projected point p on se line
* symmetrypoint(Point p) - returns reflection point of p over se
*
*/
struct Line {
Point s, e;
Line() {}
Line(Point _s, Point _e) {
s = _s;
e = _e;
}
bool operator==(Line v) {
return (s == v.s) && (e == v.e);
}
//根据一个点和倾斜角 angle 确定直线,0<=angle<pi
Line(Point p, double angle) {
s = p;
if (sgn(angle - pi / 2) == 0) {
e = (s + Point(0, 1));
} else {
e = (s + Point(1, tan(angle)));
}
}
// ax+by+c=0
Line(double a, double b, double c) {
if (sgn(a) == 0) {
s = Point(0, -c / b);
e = Point(1, -c / b);
} else if (sgn(b) == 0) {
s = Point(-c / a, 0);
e = Point(-c / a, 1);
} else {
s = Point(0, -c / b);
e = Point(1, (-c - a) / b);
}
}
void input() {
s.input();
e.input();
}
void adjust() {
if (e < s) swap(s, e);
}
//求线段长度
double length() {
return s.distance(e);
}
//返回直线倾斜角 0<=angle<pi
double angle() {
double k = atan2(e.y - s.y, e.x - s.x);
if (sgn(k) < 0) k += pi;
if (sgn(k - pi) == 0) k -= pi;
return k;
}
//点和直线关系
// 1 在左侧
// 2 在右侧
// 3 在直线上
int relation(Point p) {
int c = sgn((p - s) ^ (e - s));
if (c < 0)
return 1;
else if (c > 0)
return 2;
else
return 3;
}
// 点在线段上的判断
bool pointonseg(Point p) {
return sgn((p - s) ^ (e - s)) == 0 && sgn((p - s) * (p - e)) <= 0;
}
//两向量平行 (对应直线平行或重合)
bool parallel(Line v) {
return sgn((e - s) ^ (v.e - v.s)) == 0;
}
//两线段相交判断
// 2 规范相交
// 1 非规范相交
// 0 不相交
int segcrossseg(Line v) {
int d1 = sgn((e - s) ^ (v.s - s));
int d2 = sgn((e - s) ^ (v.e - s));
int d3 = sgn((v.e - v.s) ^ (s - v.s));
int d4 = sgn((v.e - v.s) ^ (e - v.s));
if ((d1 ^ d2) == -2 && (d3 ^ d4) == -2) return 2;
return (d1 == 0 && sgn((v.s - s) * (v.s - e)) <= 0) || (d2 == 0 && sgn((v.e - s) * (v.e - e)) <= 0) || (d3 == 0 && sgn((s - v.s) * (s - v.e)) <= 0) || (d4 == 0 && sgn((e - v.s) * (e - v.e)) <= 0);
}
//直线和线段相交判断
//-*this line -v seg
// 2 规范相交
// 1 非规范相交
// 0 不相交
int linecrossseg(Line v) {
int d1 = sgn((e - s) ^ (v.s - s));
int d2 = sgn((e - s) ^ (v.e - s));
if ((d1 ^ d2) == -2) return 2;
return (d1 == 0 || d2 == 0);
}
//两直线关系
// 0 平行
// 1 重合
// 2 相交
int linecrossline(Line v) {
if ((*this).parallel(v)) return v.relation(s) == 3;
return 2;
}
//求两直线的交点
//要保证两直线不平行或重合
Point crosspoint(Line v) {
double a1 = (v.e - v.s) ^ (s - v.s);
double a2 = (v.e - v.s) ^ (e - v.s);
return Point((s.x * a2 - e.x * a1) / (a2 - a1), (s.y * a2 - e.y * a1) / (a2 - a1));
}
//点到直线的距离
double dispointtoline(Point p) {
return fabs((p - s) ^ (e - s)) / length();
}
//点到线段的距离
double dispointtoseg(Point p) {
if (sgn((p - s) * (e - s)) < 0 || sgn((p - e) * (s - e)) < 0) return min(p.distance(s), p.distance(e));
return dispointtoline(p);
}
//返回线段到线段的距离
//前提是两线段不相交,相交距离就是 0 了
double dissegtoseg(Line v) {
return min(min(dispointtoseg(v.s), dispointtoseg(v.e)), min(v.dispointtoseg(s), v.dispointtoseg(e)));
}
//返回点 p 在直线上的投影
Point lineprog(Point p) {
return s + (((e - s) * ((e - s) * (p - s))) / ((e - s).len2()));
}
//返回点 p 关于直线的对称点
Point symmetrypoint(Point p) {
Point q = lineprog(p);
return Point(2 * q.x - p.x, 2 * q.y - p.y);
}
};
//
struct circle {
Point p; //圆心
double r; //半径
circle() {}
circle(Point _p, double _r) {
p = _p;
r = _r;
}
circle(double x, double y, double _r) {
p = Point(x, y);
r = _r;
}
//三角形的外接圆
//需要 Point 的 + / rotate() 以及 Line 的 crosspoint()
//利用两条边的中垂线得到圆心
//测试:UVA12304
circle(Point a, Point b, Point c) {
Line u = Line((a + b) / 2, ((a + b) / 2) + ((b - a).rotleft()));
Line v = Line((b + c) / 2, ((b + c) / 2) + ((c - b).rotleft()));
p = u.crosspoint(v);
r = p.distance(a);
}
//三角形的内切圆
//参数 bool t 没有作用,只是为了和上面外接圆函数区别
//测试:UVA12304
circle(Point a, Point b, Point c, bool t) {
Line u, v;
double m = atan2(b.y - a.y, b.x - a.x), n = atan2(c.y - a.y, c.x - a.x);
u.s = a;
u.e = u.s + Point(cos((n + m) / 2), sin((n + m) / 2));
v.s = b;
m = atan2(a.y - b.y, a.x - b.x), n = atan2(c.y - b.y, c.x - b.x);
v.e = v.s + Point(cos((n + m) / 2), sin((n + m) / 2));
p = u.crosspoint(v);
r = Line(a, b).dispointtoseg(p);
}
//输入
void input() {
p.input();
scanf("%lf", &r);
}
//输出
void output() {
printf("%.2lf␣%.2lf␣%.2lf\n", p.x, p.y, r);
}
bool operator==(circle v) {
return (p == v.p) && sgn(r - v.r) == 0;
}
bool operator<(circle v) const {
return ((p < v.p) || ((p == v.p) && sgn(r - v.r) < 0));
}
//面积
double area() {
return pi * r * r;
}
//周长
double circumference() {
return 2 * pi * r;
}
//点和圆的关系
// 0 圆外
// 1 圆上
// 2 圆内
int relation(Point b) {
double dst = b.distance(p);
if (sgn(dst - r) < 0)
return 2;
else if (sgn(dst - r) == 0)
return 1;
return 0;
}
//线段和圆的关系
//比较的是圆心到线段的距离和半径的关系
int relationseg(Line v) {
double dst = v.dispointtoseg(p);
if (sgn(dst - r) < 0)
return 2;
else if (sgn(dst - r) == 0)
return 1;
return 0;
}
//直线和圆的关系
//比较的是圆心到直线的距离和半径的关系
int relationline(Line v) {
double dst = v.dispointtoline(p);
if (sgn(dst - r) < 0)
return 2;
else if (sgn(dst - r) == 0)
return 1;
return 0;
}
//两圆的关系
// 5 相离
// 4 外切
// 3 相交
// 2 内切
// 1 内含
//需要 Point 的 distance
//测试:UVA12304
int relationcircle(circle v) {
double d = p.distance(v.p);
if (sgn(d - r - v.r) > 0) return 5;
if (sgn(d - r - v.r) == 0) return 4;
double l = fabs(r - v.r);
if (sgn(d - r - v.r) < 0 && sgn(d - l) > 0) return 3;
if (sgn(d - l) == 0) return 2;
if (sgn(d - l) < 0) return 1;
return 0;
}
//求两个圆的交点,返回 0 表示没有交点,返回 1 是一个交点,2 是两个交点
//需要 relationcircle
//测试:UVA12304
int pointcrosscircle(circle v, Point &p1, Point &p2) {
int rel = relationcircle(v);
if (rel == 1 || rel == 5) return 0;
double d = p.distance(v.p);
double l = (d * d + r * r - v.r * v.r) / (2 * d);
double h = sqrt(r * r - l * l);
Point tmp = p + (v.p - p).trunc(l);
p1 = tmp + ((v.p - p).rotleft().trunc(h));
p2 = tmp + ((v.p - p).rotright().trunc(h));
if (rel == 2 || rel == 4) return 1;
return 2;
}
//求直线和圆的交点,返回交点个数
int pointcrossline(Line v, Point &p1, Point &p2) {
if (!(*this).relationline(v)) return 0;
Point a = v.lineprog(p);
double d = v.dispointtoline(p);
d = sqrt(r * r - d * d);
if (sgn(d) == 0) {
p1 = a;
p2 = a;
return 1;
}
p1 = a + (v.e - v.s).trunc(d);
p2 = a - (v.e - v.s).trunc(d);
return 2;
}
//得到过 a,b 两点,半径为 r1 的两个圆
int gercircle(Point a, Point b, double r1, circle &c1, circle &c2) {
circle x(a, r1), y(b, r1);
int t = x.pointcrosscircle(y, c1.p, c2.p);
if (!t) return 0;
c1.r = c2.r = r;
return t;
}
//得到与直线 u 相切,过点 q, 半径为 r1 的圆
//测试:UVA12304
int getcircle(Line u, Point q, double r1, circle &c1, circle &c2) {
double dis = u.dispointtoline(q);
if (sgn(dis - r1 * 2) > 0) return 0;
if (sgn(dis) == 0) {
c1.p = q + ((u.e - u.s).rotleft().trunc(r1));
c2.p = q + ((u.e - u.s).rotright().trunc(r1));
c1.r = c2.r = r1;
return 2;
}
Line u1 = Line((u.s + (u.e - u.s).rotleft().trunc(r1)), (u.e + (u.e - u.s).rotleft().trunc(r1)));
Line u2 = Line((u.s + (u.e - u.s).rotright().trunc(r1)), (u.e + (u.e - u.s).rotright().trunc(r1)));
circle cc = circle(q, r1);
Point p1, p2;
if (!cc.pointcrossline(u1, p1, p2)) cc.pointcrossline(u2, p1, p2);
c1 = circle(p1, r1);
if (p1 == p2) {
c2 = c1;
return 1;
}
c2 = circle(p2, r1);
return 2;
}
//同时与直线 u,v 相切,半径为 r1 的圆
//测试:UVA12304
int getcircle(Line u, Line v, double r1, circle &c1, circle &c2, circle &c3, circle &c4) {
if (u.parallel(v)) return 0; //两直线平行
Line u1 = Line(u.s + (u.e - u.s).rotleft().trunc(r1), u.e + (u.e - u.s).rotleft().trunc(r1));
Line u2 = Line(u.s + (u.e - u.s).rotright().trunc(r1), u.e + (u.e - u.s).rotright().trunc(r1));
Line v1 = Line(v.s + (v.e - v.s).rotleft().trunc(r1), v.e + (v.e - v.s).rotleft().trunc(r1));
Line v2 = Line(v.s + (v.e - v.s).rotright().trunc(r1), v.e + (v.e - v.s).rotright().trunc(r1));
c1.r = c2.r = c3.r = c4.r = r1;
c1.p = u1.crosspoint(v1);
c2.p = u1.crosspoint(v2);
c3.p = u2.crosspoint(v1);
c4.p = u2.crosspoint(v2);
return 4;
}
//同时与不相交圆 cx,cy 相切,半径为 r1 的圆
//测试:UVA12304
int getcircle(circle cx, circle cy, double r1, circle &c1, circle &c2) {
circle x(cx.p, r1 + cx.r), y(cy.p, r1 + cy.r);
int t = x.pointcrosscircle(y, c1.p, c2.p);
if (!t) return 0;
c1.r = c2.r = r1;
return t;
}

//过一点作圆的切线 (先判断点和圆的关系)
//测试:UVA12304
int tangentline(Point q, Line &u, Line &v) {
int x = relation(q);
if (x == 2) return 0;
if (x == 1) {
u = Line(q, q + (q - p).rotleft());
v = u;
return 1;
}
double d = p.distance(q);
double l = r * r / d;
double h = sqrt(r * r - l * l);
u = Line(q, p + ((q - p).trunc(l) + (q - p).rotleft().trunc(h)));
v = Line(q, p + ((q - p).trunc(l) + (q - p).rotright().trunc(h)));
return 2;
}
//求两圆相交的面积
double areacircle(circle v) {
int rel = relationcircle(v);
if (rel >= 4) return 0.0;
if (rel <= 2) return min(area(), v.area());
double d = p.distance(v.p);
double hf = (r + v.r + d) / 2.0;
double ss = 2 * sqrt(hf * (hf - r) * (hf - v.r) * (hf - d));
double a1 = acos((r * r + d * d - v.r * v.r) / (2.0 * r * d));
a1 = a1 * r * r;
double a2 = acos((v.r * v.r + d * d - r * r) / (2.0 * v.r * d));
a2 = a2 * v.r * v.r;
return a1 + a2 - ss;
}
//求圆和三角形 pab 的相交面积
//测试:POJ3675 HDU3982 HDU2892
double areatriangle(Point a, Point b) {
if (sgn((p - a) ^ (p - b)) == 0) return 0.0;
Point q[5];
int len = 0;
q[len++] = a;
Line l(a, b);
Point p1, p2;
if (pointcrossline(l, q[1], q[2]) == 2) {
if (sgn((a - q[1]) * (b - q[1])) < 0) q[len++] = q[1];
if (sgn((a - q[2]) * (b - q[2])) < 0) q[len++] = q[2];
}
q[len++] = b;
if (len == 4 && sgn((q[0] - q[1]) * (q[2] - q[1])) > 0) swap(q[1], q[2]);
double res = 0;
for (int i = 0; i < len - 1; i++) {
if (relation(q[i]) == 0 || relation(q[i + 1]) == 0) {
double arg = p.rad(q[i], q[i + 1]);
res += r * r * arg / 2.0;
} else {
res += fabs((q[i] - p) ^ (q[i + 1] - p)) / 2.0;
}
}
return res;
}
};

/*
* n,p Line l for each side
* input(int _n) - inputs _n size polygon
* add(Point q) - adds a point at end of the list
* getline() - populates line array
* cmp - comparision in convex_hull order
* norm() - sorting in convex_hull order
* getconvex(polygon &convex) - returns convex hull in convex
* Graham(polygon &convex) - returns convex hull in convex
* isconvex() - checks if convex
* relationpoint(Point q) - returns 3 if q is a vertex
* 2 if on a side
* 1 if inside
* 0 if outside
* convexcut(Line u,polygon &po) - left side of u in po
* gercircumference() - returns side length
* getarea() - returns area
* getdir() - returns 0 for cw, 1 for ccw
* getbarycentre() - returns barycenter
*
*/

struct polygon {
int n;
Point p[maxp];
Line l[maxp];
void input(int _n) {
n = _n;
for (int i = 0; i < n; i++) p[i].input();
}
void add(Point q) {
p[n++] = q;
}
void getline() {
for (int i = 0; i < n; i++) {
l[i] = Line(p[i], p[(i + 1) % n]);
}
}
struct cmp {
Point p;
cmp(const Point &p0) {
p = p0;
}
bool operator()(const Point &aa, const Point &bb) {
Point a = aa, b = bb;
int d = sgn((a - p) ^ (b - p));
if (d == 0) {
return sgn(a.distance(p) - b.distance(p)) < 0;
}
return d > 0;
}
};
//进行极角排序
//首先需要找到最左下角的点
//需要重载号好 Point 的 < 操作符 (min 函数要用)
void norm() {
Point mi = p[0];
for (int i = 1; i < n; i++) mi = min(mi, p[i]);
sort(p, p + n, cmp(mi));
}
//得到凸包
//得到的凸包里面的点编号是 0 ∼ n-1 的
//两种凸包的方法
//注意如果有影响,要特判下所有点共点,或者共线的特殊情况
//测试 LightOJ1203 LightOJ1239
void getconvex(polygon &convex) {
sort(p, p + n);
convex.n = n;
for (int i = 0; i < min(n, 2); i++) {
convex.p[i] = p[i];
}
if (convex.n == 2 && (convex.p[0] == convex.p[1])) convex.n--; //特判
if (n <= 2) return;
int &top = convex.n;
top = 1;
for (int i = 2; i < n; i++) {
while (top && sgn((convex.p[top] - p[i]) ^ (convex.p[top - 1] - p[i])) <= 0) top--;
convex.p[++top] = p[i];
}
int temp = top;
convex.p[++top] = p[n - 2];
for (int i = n - 3; i >= 0; i--) {
while (top != temp && sgn((convex.p[top] - p[i]) ^ (convex.p[top - 1] - p[i])) <= 0) top--;
convex.p[++top] = p[i];
}
if (convex.n == 2 && (convex.p[0] == convex.p[1])) convex.n--; //特判
convex.norm(); //原来得到的是顺时针的点,排序后逆时针
}
//得到凸包的另外一种方法
//测试 LightOJ1203 LightOJ1239
void Graham(polygon &convex) {
norm();
int &top = convex.n;
top = 0;
if (n == 1) {
top = 1;
convex.p[0] = p[0];
return;
}
if (n == 2) {
top = 2;
convex.p[0] = p[0];
convex.p[1] = p[1];
if (convex.p[0] == convex.p[1]) top--;
return;
}
convex.p[0] = p[0];
convex.p[1] = p[1];
top = 2;
for (int i = 2; i < n; i++) {
while (top > 1 && sgn((convex.p[top - 1] - convex.p[top - 2]) ^ (p[i] - convex.p[top - 2])) <= 0) top--;
convex.p[top++] = p[i];
}
if (convex.n == 2 && (convex.p[0] == convex.p[1])) convex.n--; //特判
}
//判断是不是凸的
bool isconvex() {
bool s[2];
memset(s, false, sizeof(s));
for (int i = 0; i < n; i++) {
int j = (i + 1) % n;
int k = (j + 1) % n;
s[sgn((p[j] - p[i]) ^ (p[k] - p[i])) + 1] = true;
if (s[0] && s[2]) return false;
}
return true;
}
//判断点和任意多边形的关系
// 3 点上
// 2 边上
// 1 内部
// 0 外部
int relationpoint(Point q) {
for (int i = 0; i < n; i++) {
if (p[i] == q) return 3;
}
getline();
for (int i = 0; i < n; i++) {
if (l[i].pointonseg(q)) return 2;
}
int cnt = 0;
for (int i = 0; i < n; i++) {
int j = (i + 1) % n;
int k = sgn((q - p[j]) ^ (p[i] - p[j]));
int u = sgn(p[i].y - q.y);
int v = sgn(p[j].y - q.y);
if (k > 0 && u < 0 && v >= 0) cnt++;
if (k < 0 && v < 0 && u >= 0) cnt--;
}
return cnt != 0;
}
//直线 u 切割凸多边形左侧
//注意直线方向
//测试:HDU3982
void convexcut(Line u, polygon &po) {
int &top = po.n; //注意引用
top = 0;
for (int i = 0; i < n; i++) {
int d1 = sgn((u.e - u.s) ^ (p[i] - u.s));
int d2 = sgn((u.e - u.s) ^ (p[(i + 1) % n] - u.s));
if (d1 >= 0) po.p[top++] = p[i];
if (d1 * d2 < 0) po.p[top++] = u.crosspoint(Line(p[i], p[(i + 1) % n]));
}
}
//得到周长
//测试 LightOJ1239
double getcircumference() {
double sum = 0;
for (int i = 0; i < n; i++) {
sum += p[i].distance(p[(i + 1) % n]);
}
return sum;
}
//得到面积面积
double getarea() {
double sum = 0;
for (int i = 0; i < n; i++) {
sum += (p[i] ^ p[(i + 1) % n]);
}
return fabs(sum) / 2;
}
//得到方向
// 1 表示逆时针,0 表示顺时针
bool getdir() {
double sum = 0;
for (int i = 0; i < n; i++) sum += (p[i] ^ p[(i + 1) % n]);
if (sgn(sum) > 0) return 1;
return 0;
}
//得到重心
Point getbarycentre() {
Point ret(0, 0);
double area = 0;
for (int i = 1; i < n - 1; i++) {
double tmp = (p[i] - p[0]) ^ (p[i + 1] - p[0]);
if (sgn(tmp) == 0) continue;
area += tmp;
ret.x += (p[0].x + p[i].x + p[i + 1].x) / 3 * tmp;
ret.y += (p[0].y + p[i].y + p[i + 1].y) / 3 * tmp;
}
if (sgn(area)) ret = ret / area;
return ret;
}
//多边形和圆交的面积
//测试:POJ3675 HDU3982 HDU2892
double areacircle(circle c) {
double ans = 0;
for (int i = 0; i < n; i++) {
int j = (i + 1) % n;
if (sgn((p[j] - c.p) ^ (p[i] - c.p)) >= 0)
ans += c.areatriangle(p[i], p[j]);
else
ans -= c.areatriangle(p[i], p[j]);
}
return fabs(ans);
}
//多边形和圆关系
// 2 圆完全在多边形内
// 1 圆在多边形里面,碰到了多边形边界
// 0 其它
int relationcircle(circle c) {
getline();
int x = 2;
if (relationpoint(c.p) != 1) return 0; //圆心不在内部
for (int i = 0; i < n; i++) {
if (c.relationseg(l[i]) == 2) return 0;
if (c.relationseg(l[i]) == 1) x = 1;
}
return x;
}
};
// AB X AC
double cross(Point A, Point B, Point C) {
return (B - A) ^ (C - A);
}
// AB*AC
double dot(Point A, Point B, Point C) {
return (B - A) * (C - A);
}
//最小矩形面积覆盖
// A 必须是凸包 (而且是逆时针顺序)
// 测试 UVA 10173
double minRectangleCover(polygon A) {
//要特判 A.n < 3 的情况
if (A.n < 3) return 0.0;
A.p[A.n] = A.p[0];
double ans = -1;
int r = 1, p = 1, q;
for (int i = 0; i < A.n; i++) {
//卡出离边 A.p[i] - A.p[i+1] 最远的点
while (sgn(cross(A.p[i], A.p[i + 1], A.p[r + 1]) - cross(A.p[i], A.p[i + 1], A.p[r])) >= 0) r = (r + 1) % A.n;
//卡出 A.p[i] - A.p[i+1] 方向上正向 n 最远的点
while (sgn(dot(A.p[i], A.p[i + 1], A.p[p + 1]) - dot(A.p[i], A.p[i + 1], A.p[p])) >= 0) p = (p + 1) % A.n;
if (i == 0) q = p;
//卡出 A.p[i] - A.p[i+1] 方向上负向最远的点
while (sgn(dot(A.p[i], A.p[i + 1], A.p[q + 1]) - dot(A.p[i], A.p[i + 1], A.p[q])) <= 0) q = (q + 1) % A.n;
double d = (A.p[i] - A.p[i + 1]).len2();
double tmp = cross(A.p[i], A.p[i + 1], A.p[r]) * (dot(A.p[i], A.p[i + 1], A.p[p]) - dot(A.p[i], A.p[i + 1], A.p[q])) / d;
if (ans < 0 || ans > tmp) ans = tmp;
}
return ans;
}

//直线切凸多边形
//多边形是逆时针的,在 q1q2 的左侧
//测试:HDU3982
vector<Point> convexCut(const vector<Point> &ps, Point q1, Point q2) {
vector<Point> qs;
int n = ps.size();
for (int i = 0; i < n; i++) {
Point p1 = ps[i], p2 = ps[(i + 1) % n];
int d1 = sgn((q2 - q1) ^ (p1 - q1)), d2 = sgn((q2 - q1) ^ (p2 - q1));
if (d1 >= 0) qs.push_back(p1);
if (d1 * d2 < 0) qs.push_back(Line(p1, p2).crosspoint(Line(q1, q2)));
}
return qs;
}
double l = 0.00001, r = 3000;
double cx, cy;
polygon P, P2;
double p, q;
double area;
void slove() {
scanf("%lf%lf", &cx, &cy);
scanf("%lf%lf", &p, &q);
double a = area * (q - p) / q;
// cout << "a==" << a << endl;

l = 0.00001, r = 3000;
while (l < r) {
double mid = (l + r) / 2.0;
circle c(cx, cy, mid);
double res = P.areacircle(c);
if (fabs(res - a) < eps) {
printf("%.8f\n", mid);
return;
} else if (res < a) {
l = mid;
} else {
r = mid;
}
}
}

const int maxn = 1e8;
const int mod = 32767;
int main() {
int n, m;
scanf("%d", &n);
P.input(n);
area = P.getarea();
scanf("%d", &m);
// cout << "area==" << area << endl;
for (int i = 0; i < m; ++i) {
slove();
}
}

回文树学习笔记

回文树学习笔记

神奇的数据结构 Palindromic Tree 回文树

功能简介

  1. 求一个串$S$中$[0,i]$中本质不同回文串的个数
  2. 求串$S$中每一个本质不同回文串出现的次数
  3. 求指定下标$i$结尾的回文串的个数

变量简介

  1. $len[i]$表示编号为$i$节点所表示的回文串的$len$。
  2. $next[i][c]$表示编号为$i$的节点表示的回文串在两边添加字符$c$之后会变成的回文串的编号。
  3. $fail[i]$表示节点$i$失配以后跳转后不等于自己的节点$i$所表示的回稳产的最长长度。
  4. $cnt[i]$表示节点$i$表示的本质不同的串的个数。
  5. $num[i]$表示以节点$i$表示的最长回文串的最右端点为回文串结尾的回文串个数。
  6. $last$指向新添加一个字母后所形成的最长回文串表示的节点。
  7. $S[i]$表示第$i$次添加的字符(初始化S[0]=不存在的字符)。
  8. $p$表示添加的节点个数。
  9. $n$表示添加的字符个数。

模板(抄的)

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const int maxn = 1e5+10;
const int N = 26; // attention 用模板的时候要改变

struct Ptree{
int next[maxn][N];
int fail[maxn];
int cnt[maxn];
int num[maxn];
int len[maxn];
int S[maxn];
int last,n,p;
int newNode(int x){
rep(i,0,N) next[p][i] = 0;
cnt[p] = 0;num[p] = 0;len[p] = x;
return p++;
}
void init(){
p = 0;
newNode(0);newNode(-1);
last = 0; n = 0;S[n] = -1;fail[0] = 1;
}
int get_fail(int x){
while(S[n-len[x]-1] != S[n]) x = fail[x];
return x;
}
void add(int c){
c -= 'a'; // attention 用模板的时候要改变
S[++n] = c;
int cur = get_fail(last);
if(!next[cur][c]){
int now = newNode(len[cur]+2);
fail[now] = next[get_fail(fail[cur])][c];
next[cur][c] = now;
num[now] = num[fail[now]] + 1;
}
last = next[cur][c];
cnt[last]++;
}
void count(){
per(i,0,p) cnt[fail[i]] += cnt[i];
}
}

https://blog.csdn.net/u013368721/article/details/42100363