计算几何全家桶

#include <bits/stdc++.h>
using namespace std;using point_t=long double;  //全局数据类型，可修改为 long long 等constexpr point_t eps=1e-8;
constexpr long double PI=3.1415926535897932384l;// 点与向量
template<typename T> struct point
{T x,y;bool operator==(const point &a) const {return (abs(x-a.x)<=eps && abs(y-a.y)<=eps);}bool operator<(const point &a) const {if (abs(x-a.x)<=eps) return y<a.y-eps; return x<a.x-eps;}bool operator>(const point &a) const {return !(*this<a || *this==a);}point operator+(const point &a) const {return {x+a.x,y+a.y};}point operator-(const point &a) const {return {x-a.x,y-a.y};}point operator-() const {return {-x,-y};}point operator*(const T k) const {return {k*x,k*y};}point operator/(const T k) const {return {x/k,y/k};}T operator*(const point &a) const {return x*a.x+y*a.y;}  // 点积T operator^(const point &a) const {return x*a.y-y*a.x;}  // 叉积，注意优先级int toleft(const point &a) const {const auto t=(*this)^a; return (t>eps)-(t<-eps);}  // to-left 测试T len2() const {return (*this)*(*this);}  // 向量长度的平方T dis2(const point &a) const {return (a-(*this)).len2();}  // 两点距离的平方// 涉及浮点数long double len() const {return sqrtl(len2());}  // 向量长度long double dis(const point &a) const {return sqrtl(dis2(a));}  // 两点距离long double ang(const point &a) const {return acosl(max(-1.0l,min(1.0l,((*this)*a)/(len()*a.len()))));}  // 向量夹角point rot(const long double rad) const {return {x*cos(rad)-y*sin(rad),x*sin(rad)+y*cos(rad)};}  // 逆时针旋转（给定角度）point rot(const long double cosr,const long double sinr) const {return {x*cosr-y*sinr,x*sinr+y*cosr};}  // 逆时针旋转（给定角度的正弦与余弦）
};using Point=point<point_t>;// 极角排序
struct argcmp
{bool operator()(const Point &a,const Point &b) const{const auto quad=[](const Point &a){if (a.y<-eps) return 1;if (a.y>eps) return 4;if (a.x<-eps) return 5;if (a.x>eps) return 3;return 2;};const int qa=quad(a),qb=quad(b);if (qa!=qb) return qa<qb;const auto t=a^b;// if (abs(t)<=eps) return a*a<b*b-eps;  // 不同长度的向量需要分开return t>eps;}
};// 直线
template<typename T> struct line
{point<T> p,v;  // p 为直线上一点，v 为方向向量bool operator==(const line &a) const {return v.toleft(a.v)==0 && v.toleft(p-a.p)==0;}int toleft(const point<T> &a) const {return v.toleft(a-p);}  // to-left 测试bool operator<(const line &a) const  // 半平面交算法定义的排序{if (abs(v^a.v)<=eps && v*a.v>=-eps) return toleft(a.p)==-1;return argcmp()(v,a.v);}// 涉及浮点数point<T> inter(const line &a) const {return p+v*((a.v^(p-a.p))/(v^a.v));}  // 直线交点long double dis(const point<T> &a) const {return abs(v^(a-p))/v.len();}  // 点到直线距离point<T> proj(const point<T> &a) const {return p+v*((v*(a-p))/(v*v));}  // 点在直线上的投影
};using Line=line<point_t>;//线段
template<typename T> struct segment
{point<T> a,b;bool operator<(const segment &s) const {return make_pair(a,b)<make_pair(s.a,s.b);}// 判定性函数建议在整数域使用// 判断点是否在线段上// -1 点在线段端点 | 0 点不在线段上 | 1 点严格在线段上int is_on(const point<T> &p) const  {if (p==a || p==b) return -1;return (p-a).toleft(p-b)==0 && (p-a)*(p-b)<-eps;}// 判断线段直线是否相交// -1 直线经过线段端点 | 0 线段和直线不相交 | 1 线段和直线严格相交int is_inter(const line<T> &l) const{if (l.toleft(a)==0 || l.toleft(b)==0) return -1;return l.toleft(a)!=l.toleft(b);}// 判断两线段是否相交// -1 在某一线段端点处相交 | 0 两线段不相交 | 1 两线段严格相交int is_inter(const segment<T> &s) const{if (is_on(s.a) || is_on(s.b) || s.is_on(a) || s.is_on(b)) return -1;const line<T> l{a,b-a},ls{s.a,s.b-s.a};return l.toleft(s.a)*l.toleft(s.b)==-1 && ls.toleft(a)*ls.toleft(b)==-1;}// 点到线段距离long double dis(const point<T> &p) const{if ((p-a)*(b-a)<-eps || (p-b)*(a-b)<-eps) return min(p.dis(a),p.dis(b));const line<T> l{a,b-a};return l.dis(p);}// 两线段间距离long double dis(const segment<T> &s) const{if (is_inter(s)) return 0;return min({dis(s.a),dis(s.b),s.dis(a),s.dis(b)});}
};using Segment=segment<point_t>;// 多边形
template<typename T> struct polygon
{vector<point<T>> p;  // 以逆时针顺序存储size_t nxt(const size_t i) const {return i==p.size()-1?0:i+1;}size_t pre(const size_t i) const {return i==0?p.size()-1:i-1;}// 回转数// 返回值第一项表示点是否在多边形边上// 对于狭义多边形，回转数为 0 表示点在多边形外，否则点在多边形内pair<bool,int> winding(const point<T> &a) const{int cnt=0;for (size_t i=0;i<p.size();i++){const point<T> u=p[i],v=p[nxt(i)];if (abs((a-u)^(a-v))<=eps && (a-u)*(a-v)<=eps) return {true,0};if (abs(u.y-v.y)<=eps) continue;const Line uv={u,v-u};if (u.y<v.y-eps && uv.toleft(a)<=0) continue;if (u.y>v.y+eps && uv.toleft(a)>=0) continue;if (u.y<a.y-eps && v.y>=a.y-eps) cnt++;if (u.y>=a.y-eps && v.y<a.y-eps) cnt--;}return {false,cnt};}// 多边形面积的两倍// 可用于判断点的存储顺序是顺时针或逆时针T area() const{T sum=0;for (size_t i=0;i<p.size();i++) sum+=p[i]^p[nxt(i)];return sum;}// 多边形的周长long double circ() const{long double sum=0;for (size_t i=0;i<p.size();i++) sum+=p[i].dis(p[nxt(i)]);return sum;}
};using Polygon=polygon<point_t>;//凸多边形
template<typename T> struct convex: polygon<T>
{// 闵可夫斯基和convex operator+(const convex &c) const  {const auto &p=this->p;vector<Segment> e1(p.size()),e2(c.p.size()),edge(p.size()+c.p.size());vector<point<T>> res; res.reserve(p.size()+c.p.size());const auto cmp=[](const Segment &u,const Segment &v) {return argcmp()(u.b-u.a,v.b-v.a);};for (size_t i=0;i<p.size();i++) e1[i]={p[i],p[this->nxt(i)]};for (size_t i=0;i<c.p.size();i++) e2[i]={c.p[i],c.p[c.nxt(i)]};rotate(e1.begin(),min_element(e1.begin(),e1.end(),cmp),e1.end());rotate(e2.begin(),min_element(e2.begin(),e2.end(),cmp),e2.end());merge(e1.begin(),e1.end(),e2.begin(),e2.end(),edge.begin(),cmp);const auto check=[](const vector<point<T>> &res,const point<T> &u){const auto back1=res.back(),back2=*prev(res.end(),2);return (back1-back2).toleft(u-back1)==0 && (back1-back2)*(u-back1)>=-eps;};auto u=e1[0].a+e2[0].a;for (const auto &v:edge){while (res.size()>1 && check(res,u)) res.pop_back();res.push_back(u);u=u+v.b-v.a;}if (res.size()>1 && check(res,res[0])) res.pop_back();return {res};}// 旋转卡壳// func 为更新答案的函数，可以根据题目调整位置template<typename F> void rotcaliper(const F &func) const{const auto &p=this->p;const auto area=[](const point<T> &u,const point<T> &v,const point<T> &w){return (w-u)^(w-v);};for (size_t i=0,j=1;i<p.size();i++){const auto nxti=this->nxt(i);func(p[i],p[nxti],p[j]);while (area(p[this->nxt(j)],p[i],p[nxti])>=area(p[j],p[i],p[nxti])){j=this->nxt(j);func(p[i],p[nxti],p[j]);}}}// 凸多边形的直径的平方T diameter2() const{const auto &p=this->p;if (p.size()==1) return 0;if (p.size()==2) return p[0].dis2(p[1]);T ans=0;auto func=[&](const point<T> &u,const point<T> &v,const point<T> &w){ans=max({ans,w.dis2(u),w.dis2(v)});};rotcaliper(func);return ans;}// 判断点是否在凸多边形内// 复杂度 O(logn)// -1 点在多边形边上 | 0 点在多边形外 | 1 点在多边形内int is_in(const point<T> &a) const{const auto &p=this->p;if (p.size()==1) return a==p[0]?-1:0;if (p.size()==2) return segment<T>{p[0],p[1]}.is_on(a)?-1:0; if (a==p[0]) return -1;if ((p[1]-p[0]).toleft(a-p[0])==-1 || (p.back()-p[0]).toleft(a-p[0])==1) return 0;const auto cmp=[&](const Point &u,const Point &v){return (u-p[0]).toleft(v-p[0])==1;};const size_t i=lower_bound(p.begin()+1,p.end(),a,cmp)-p.begin();if (i==1) return segment<T>{p[0],p[i]}.is_on(a)?-1:0;if (i==p.size()-1 && segment<T>{p[0],p[i]}.is_on(a)) return -1;if (segment<T>{p[i-1],p[i]}.is_on(a)) return -1;return (p[i]-p[i-1]).toleft(a-p[i-1])>0;}// 凸多边形关于某一方向的极点// 复杂度 O(logn)// 参考资料：https://codeforces.com/blog/entry/48868template<typename F> size_t extreme(const F &dir) const{const auto &p=this->p;const auto check=[&](const size_t i){return dir(p[i]).toleft(p[this->nxt(i)]-p[i])>=0;};const auto dir0=dir(p[0]); const auto check0=check(0);if (!check0 && check(p.size()-1)) return 0;const auto cmp=[&](const Point &v){const size_t vi=&v-p.data();if (vi==0) return 1;const auto checkv=check(vi);const auto t=dir0.toleft(v-p[0]);if (vi==1 && checkv==check0 && t==0) return 1;return checkv^(checkv==check0 && t<=0);};return partition_point(p.begin(),p.end(),cmp)-p.begin();}// 过凸多边形外一点求凸多边形的切线，返回切点下标// 复杂度 O(logn)// 必须保证点在多边形外pair<size_t,size_t> tangent(const point<T> &a) const{const size_t i=extreme([&](const point<T> &u){return u-a;});const size_t j=extreme([&](const point<T> &u){return a-u;});return {i,j};}// 求平行于给定直线的凸多边形的切线，返回切点下标// 复杂度 O(logn)pair<size_t,size_t> tangent(const line<T> &a) const{const size_t i=extreme([&](...){return a.v;});const size_t j=extreme([&](...){return -a.v;});return {i,j};}
};using Convex=convex<point_t>;// 圆
struct Circle
{Point c;long double r;bool operator==(const Circle &a) const {return c==a.c && abs(r-a.r)<=eps;}long double circ() const {return 2*PI*r;}  // 周长long double area() const {return PI*r*r;}  // 面积// 点与圆的关系// -1 圆上 | 0 圆外 | 1 圆内int is_in(const Point &p) const {const long double d=p.dis(c); return abs(d-r)<=eps?-1:d<r-eps;}// 直线与圆关系// 0 相离 | 1 相切 | 2 相交int relation(const Line &l) const{const long double d=l.dis(c);if (d>r+eps) return 0;if (abs(d-r)<=eps) return 1;return 2;}// 圆与圆关系// -1 相同 | 0 相离 | 1 外切 | 2 相交 | 3 内切 | 4 内含int relation(const Circle &a) const{if (*this==a) return -1;const long double d=c.dis(a.c);if (d>r+a.r+eps) return 0;if (abs(d-r-a.r)<=eps) return 1;if (abs(d-abs(r-a.r))<=eps) return 3;if (d<abs(r-a.r)-eps) return 4;return 2;}// 直线与圆的交点vector<Point> inter(const Line &l) const{const long double d=l.dis(c);const Point p=l.proj(c);const int t=relation(l);if (t==0) return vector<Point>();if (t==1) return vector<Point>{p};const long double k=sqrt(r*r-d*d);return vector<Point>{p-(l.v/l.v.len())*k,p+(l.v/l.v.len())*k};}// 圆与圆交点vector<Point> inter(const Circle &a) const{const long double d=c.dis(a.c);const int t=relation(a);if (t==-1 || t==0 || t==4) return vector<Point>();Point e=a.c-c; e=e/e.len()*r;if (t==1 || t==3) {if (r*r+d*d-a.r*a.r>=-eps) return vector<Point>{c+e};return vector<Point>{c-e};}const long double costh=(r*r+d*d-a.r*a.r)/(2*r*d),sinth=sqrt(1-costh*costh);return vector<Point>{c+e.rot(costh,-sinth),c+e.rot(costh,sinth)};}// 圆与圆交面积long double inter_area(const Circle &a) const{const long double d=c.dis(a.c);const int t=relation(a);if (t==-1) return area();if (t<2) return 0;if (t>2) return min(area(),a.area());const long double costh1=(r*r+d*d-a.r*a.r)/(2*r*d),costh2=(a.r*a.r+d*d-r*r)/(2*a.r*d);const long double sinth1=sqrt(1-costh1*costh1),sinth2=sqrt(1-costh2*costh2);const long double th1=acos(costh1),th2=acos(costh2);return r*r*(th1-costh1*sinth1)+a.r*a.r*(th2-costh2*sinth2);}// 过圆外一点圆的切线vector<Line> tangent(const Point &a) const{const int t=is_in(a);if (t==1) return vector<Line>();if (t==-1){const Point v={-(a-c).y,(a-c).x};return vector<Line>{{a,v}};}Point e=a-c; e=e/e.len()*r;const long double costh=r/c.dis(a),sinth=sqrt(1-costh*costh);const Point t1=c+e.rot(costh,-sinth),t2=c+e.rot(costh,sinth);return vector<Line>{{a,t1-a},{a,t2-a}};}// 两圆的公切线vector<Line> tangent(const Circle &a) const{const int t=relation(a);vector<Line> lines;if (t==-1 || t==4) return lines;if (t==1 || t==3){const Point p=inter(a)[0],v={-(a.c-c).y,(a.c-c).x};lines.push_back({p,v});}const long double d=c.dis(a.c);const Point e=(a.c-c)/(a.c-c).len();if (t<=2){const long double costh=(r-a.r)/d,sinth=sqrt(1-costh*costh);const Point d1=e.rot(costh,-sinth),d2=e.rot(costh,sinth);const Point u1=c+d1*r,u2=c+d2*r,v1=a.c+d1*a.r,v2=a.c+d2*a.r;lines.push_back({u1,v1-u1}); lines.push_back({u2,v2-u2});}if (t==0){const long double costh=(r+a.r)/d,sinth=sqrt(1-costh*costh);const Point d1=e.rot(costh,-sinth),d2=e.rot(costh,sinth);const Point u1=c+d1*r,u2=c+d2*r,v1=a.c-d1*a.r,v2=a.c-d2*a.r;lines.push_back({u1,v1-u1}); lines.push_back({u2,v2-u2});}return lines;}// 圆的反演tuple<int,Circle,Line> inverse(const Line &l) const{const Circle null_c={{0.0,0.0},0.0};const Line null_l={{0.0,0.0},{0.0,0.0}};if (l.toleft(c)==0) return {2,null_c,l};const Point v=l.toleft(c)==1?Point{l.v.y,-l.v.x}:Point{-l.v.y,l.v.x};const long double d=r*r/l.dis(c);const Point p=c+v/v.len()*d;return {1,{(c+p)/2,d/2},null_l};}tuple<int,Circle,Line> inverse(const Circle &a) const{const Circle null_c={{0.0,0.0},0.0};const Line null_l={{0.0,0.0},{0.0,0.0}};const Point v=a.c-c;if (a.is_in(c)==-1){const long double d=r*r/(a.r+a.r);const Point p=c+v/v.len()*d;return {2,null_c,{p,{-v.y,v.x}}};}if (c==a.c) return {1,{c,r*r/a.r},null_l};const long double d1=r*r/(c.dis(a.c)-a.r),d2=r*r/(c.dis(a.c)+a.r);const Point p=c+v/v.len()*d1,q=c+v/v.len()*d2;return {1,{(p+q)/2,p.dis(q)/2},null_l};}
};// 圆与多边形面积交
long double area_inter(const Circle &circ,const Polygon &poly)
{const auto cal=[](const Circle &circ,const Point &a,const Point &b){if ((a-circ.c).toleft(b-circ.c)==0) return 0.0l;const auto ina=circ.is_in(a),inb=circ.is_in(b);const Line ab={a,b-a};if (ina && inb) return ((a-circ.c)^(b-circ.c))/2;if (ina && !inb){const auto t=circ.inter(ab);const Point p=t.size()==1?t[0]:t[1];const long double ans=((a-circ.c)^(p-circ.c))/2;const long double th=(p-circ.c).ang(b-circ.c);const long double d=circ.r*circ.r*th/2;if ((a-circ.c).toleft(b-circ.c)==1) return ans+d;return ans-d;}if (!ina && inb){const Point p=circ.inter(ab)[0];const long double ans=((p-circ.c)^(b-circ.c))/2;const long double th=(a-circ.c).ang(p-circ.c);const long double d=circ.r*circ.r*th/2;if ((a-circ.c).toleft(b-circ.c)==1) return ans+d;return ans-d;}const auto p=circ.inter(ab);if (p.size()==2 && Segment{a,b}.dis(circ.c)<=circ.r+eps){const long double ans=((p[0]-circ.c)^(p[1]-circ.c))/2;const long double th1=(a-circ.c).ang(p[0]-circ.c),th2=(b-circ.c).ang(p[1]-circ.c);const long double d1=circ.r*circ.r*th1/2,d2=circ.r*circ.r*th2/2;if ((a-circ.c).toleft(b-circ.c)==1) return ans+d1+d2;return ans-d1-d2;}const long double th=(a-circ.c).ang(b-circ.c);if ((a-circ.c).toleft(b-circ.c)==1) return circ.r*circ.r*th/2;return -circ.r*circ.r*th/2;};long double ans=0;for (size_t i=0;i<poly.p.size();i++){const Point a=poly.p[i],b=poly.p[poly.nxt(i)];ans+=cal(circ,a,b);}return ans;
}// 点集的凸包
// Andrew 算法，复杂度 O(nlogn)
Convex convexhull(vector<Point> p)
{vector<Point> st;if (p.empty()) return Convex{st};sort(p.begin(),p.end());const auto check=[](const vector<Point> &st,const Point &u){const auto back1=st.back(),back2=*prev(st.end(),2);return (back1-back2).toleft(u-back1)<=0;};for (const Point &u:p){while (st.size()>1 && check(st,u)) st.pop_back();st.push_back(u);}size_t k=st.size();p.pop_back(); reverse(p.begin(),p.end());for (const Point &u:p){while (st.size()>k && check(st,u)) st.pop_back();st.push_back(u);}st.pop_back();return Convex{st};
}// 半平面交
// 排序增量法，复杂度 O(nlogn)
// 输入与返回值都是用直线表示的半平面集合
vector<Line> halfinter(vector<Line> l, const point_t lim=1e9)
{const auto check=[](const Line &a,const Line &b,const Line &c){return a.toleft(b.inter(c))<0;};// 无精度误差的方法，但注意取值范围会扩大到三次方/*const auto check=[](const Line &a,const Line &b,const Line &c){const Point p=a.v*(b.v^c.v),q=b.p*(b.v^c.v)+b.v*(c.v^(b.p-c.p))-a.p*(b.v^c.v);return p.toleft(q)<0;};*/l.push_back({{-lim,0},{0,-1}}); l.push_back({{0,-lim},{1,0}});l.push_back({{lim,0},{0,1}}); l.push_back({{0,lim},{-1,0}});sort(l.begin(),l.end());deque<Line> q;for (size_t i=0;i<l.size();i++){if (i>0 && l[i-1].v.toleft(l[i].v)==0 && l[i-1].v*l[i].v>eps) continue;while (q.size()>1 && check(l[i],q.back(),q[q.size()-2])) q.pop_back();while (q.size()>1 && check(l[i],q[0],q[1])) q.pop_front();if (!q.empty() && q.back().v.toleft(l[i].v)<=0) return vector<Line>();q.push_back(l[i]);}while (q.size()>1 && check(q[0],q.back(),q[q.size()-2])) q.pop_back();while (q.size()>1 && check(q.back(),q[0],q[1])) q.pop_front();return vector<Line>(q.begin(),q.end());
}// 点集形成的最小最大三角形
// 极角序扫描线，复杂度 O(n^2logn)
// 最大三角形问题可以使用凸包与旋转卡壳做到 O(n^2)
pair<point_t,point_t> minmax_triangle(const vector<Point> &vec)
{if (vec.size()<=2) return {0,0};vector<pair<int,int>> evt;evt.reserve(vec.size()*vec.size());point_t maxans=0,minans=numeric_limits<point_t>::max();for (size_t i=0;i<vec.size();i++){for (size_t j=0;j<vec.size();j++){if (i==j) continue;if (vec[i]==vec[j]) minans=0;else evt.push_back({i,j});}}sort(evt.begin(),evt.end(),[&](const pair<int,int> &u,const pair<int,int> &v){const Point du=vec[u.second]-vec[u.first],dv=vec[v.second]-vec[v.first];return argcmp()({du.y,-du.x},{dv.y,-dv.x});});vector<size_t> vx(vec.size()),pos(vec.size());for (size_t i=0;i<vec.size();i++) vx[i]=i;sort(vx.begin(),vx.end(),[&](int x,int y){return vec[x]<vec[y];});for (size_t i=0;i<vx.size();i++) pos[vx[i]]=i;for (auto [u,v]:evt){const size_t i=pos[u],j=pos[v];const size_t l=min(i,j),r=max(i,j);const Point vecu=vec[u],vecv=vec[v];if (l>0) minans=min(minans,abs((vec[vx[l-1]]-vecu)^(vec[vx[l-1]]-vecv)));if (r<vx.size()-1) minans=min(minans,abs((vec[vx[r+1]]-vecu)^(vec[vx[r+1]]-vecv)));maxans=max({maxans,abs((vec[vx[0]]-vecu)^(vec[vx[0]]-vecv)),abs((vec[vx.back()]-vecu)^(vec[vx.back()]-vecv))});if (i<j) swap(vx[i],vx[j]),pos[u]=j,pos[v]=i;}return {minans,maxans};
}// 判断多条线段是否有交点
// 扫描线，复杂度 O(nlogn)
bool segs_inter(const vector<Segment> &segs)
{if (segs.empty()) return false;using seq_t=tuple<point_t,int,Segment>;const auto seqcmp=[](const seq_t &u, const seq_t &v){const auto [u0,u1,u2]=u;const auto [v0,v1,v2]=v;if (abs(u0-v0)<=eps) return make_pair(u1,u2)<make_pair(v1,v2);return u0<v0-eps;};vector<seq_t> seq;for (auto seg:segs){if (seg.a.x>seg.b.x+eps) swap(seg.a,seg.b);seq.push_back({seg.a.x,0,seg});seq.push_back({seg.b.x,1,seg});}sort(seq.begin(),seq.end(),seqcmp);point_t x_now;auto cmp=[&](const Segment &u, const Segment &v){if (abs(u.a.x-u.b.x)<=eps || abs(v.a.x-v.b.x)<=eps) return u.a.y<v.a.y-eps;return ((x_now-u.a.x)*(u.b.y-u.a.y)+u.a.y*(u.b.x-u.a.x))*(v.b.x-v.a.x)<((x_now-v.a.x)*(v.b.y-v.a.y)+v.a.y*(v.b.x-v.a.x))*(u.b.x-u.a.x)-eps;};multiset<Segment,decltype(cmp)> s{cmp};for (const auto [x,o,seg]:seq){x_now=x;const auto it=s.lower_bound(seg);if (o==0){if (it!=s.end() && seg.is_inter(*it)) return true;if (it!=s.begin() && seg.is_inter(*prev(it))) return true;s.insert(seg);}else{if (next(it)!=s.end() && it!=s.begin() && (*prev(it)).is_inter(*next(it))) return true;s.erase(it);}}return false;
}// 多边形面积并
// 轮廓积分，复杂度 O(n^2logn)，n为边数
// ans[i] 表示被至少覆盖了 i+1 次的区域的面积
vector<long double> area_union(const vector<Polygon> &polys)
{const size_t siz=polys.size();vector<vector<pair<Point,Point>>> segs(siz);const auto check=[](const Point &u,const Segment &e){return !((u<e.a && u<e.b) || (u>e.a && u>e.b));};auto cut_edge=[&](const Segment &e,const size_t i){const Line le{e.a,e.b-e.a};vector<pair<Point,int>> evt;evt.push_back({e.a,0}); evt.push_back({e.b,0});for (size_t j=0;j<polys.size();j++){if (i==j) continue;const auto &pj=polys[j];for (size_t k=0;k<pj.p.size();k++){const Segment s={pj.p[k],pj.p[pj.nxt(k)]};if (le.toleft(s.a)==0 && le.toleft(s.b)==0){evt.push_back({s.a,0});evt.push_back({s.b,0});}else if (s.is_inter(le)){const Line ls{s.a,s.b-s.a};const Point u=le.inter(ls);if (le.toleft(s.a)<0 && le.toleft(s.b)>=0) evt.push_back({u,-1});else if (le.toleft(s.a)>=0 && le.toleft(s.b)<0) evt.push_back({u,1});}}}sort(evt.begin(),evt.end());if (e.a>e.b) reverse(evt.begin(),evt.end());int sum=0;for (size_t i=0;i<evt.size();i++){sum+=evt[i].second;const Point u=evt[i].first,v=evt[i+1].first;if (!(u==v) && check(u,e) && check(v,e)) segs[sum].push_back({u,v});if (v==e.b) break;}};for (size_t i=0;i<polys.size();i++){const auto &pi=polys[i];for (size_t k=0;k<pi.p.size();k++){const Segment ei={pi.p[k],pi.p[pi.nxt(k)]};cut_edge(ei,i);}}vector<long double> ans(siz);for (size_t i=0;i<siz;i++){long double sum=0;sort(segs[i].begin(),segs[i].end());int cnt=0;for (size_t j=0;j<segs[i].size();j++){if (j>0 && segs[i][j]==segs[i][j-1]) segs[i+(++cnt)].push_back(segs[i][j]);else cnt=0,sum+=segs[i][j].first^segs[i][j].second;}ans[i]=sum/2;}return ans;
}// 圆面积并
// 轮廓积分，复杂度 O(n^2logn)
// ans[i] 表示被至少覆盖了 i+1 次的区域的面积
vector<long double> area_union(const vector<Circle> &circs)
{const size_t siz=circs.size();using arc_t=tuple<Point,long double,long double,long double>;vector<vector<arc_t>> arcs(siz);const auto eq=[](const arc_t &u,const arc_t &v){const auto [u1,u2,u3,u4]=u;const auto [v1,v2,v3,v4]=v;return u1==v1 && abs(u2-v2)<=eps && abs(u3-v3)<=eps && abs(u4-v4)<=eps;};auto cut_circ=[&](const Circle &ci,const size_t i){vector<pair<long double,int>> evt;evt.push_back({-PI,0}); evt.push_back({PI,0});int init=0;for (size_t j=0;j<circs.size();j++){if (i==j) continue;const Circle &cj=circs[j];if (ci.r<cj.r-eps && ci.relation(cj)>=3) init++;const auto inters=ci.inter(cj);if (inters.size()==1) evt.push_back({atan2l((inters[0]-ci.c).y,(inters[0]-ci.c).x),0});if (inters.size()==2){const Point dl=inters[0]-ci.c,dr=inters[1]-ci.c;long double argl=atan2l(dl.y,dl.x),argr=atan2l(dr.y,dr.x);if (abs(argl+PI)<=eps) argl=PI;if (abs(argr+PI)<=eps) argr=PI;if (argl>argr+eps){evt.push_back({argl,1}); evt.push_back({PI,-1});evt.push_back({-PI,1}); evt.push_back({argr,-1});}else{evt.push_back({argl,1});evt.push_back({argr,-1});}}}sort(evt.begin(),evt.end());int sum=init;for (size_t i=0;i<evt.size();i++){sum+=evt[i].second;if (abs(evt[i].first-evt[i+1].first)>eps) arcs[sum].push_back({ci.c,ci.r,evt[i].first,evt[i+1].first});if (abs(evt[i+1].first-PI)<=eps) break;}};const auto oint=[](const arc_t &arc){const auto [cc,cr,l,r]=arc;if (abs(r-l-PI-PI)<=eps) return 2.0l*PI*cr*cr;return cr*cr*(r-l)+cc.x*cr*(sin(r)-sin(l))-cc.y*cr*(cos(r)-cos(l));};for (size_t i=0;i<circs.size();i++){const auto &ci=circs[i];cut_circ(ci,i);}vector<long double> ans(siz);for (size_t i=0;i<siz;i++){long double sum=0;sort(arcs[i].begin(),arcs[i].end());int cnt=0;for (size_t j=0;j<arcs[i].size();j++){if (j>0 && eq(arcs[i][j],arcs[i][j-1])) arcs[i+(++cnt)].push_back(arcs[i][j]);else cnt=0,sum+=oint(arcs[i][j]);}ans[i]=sum/2;}return ans;
}