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();
}
}