BZOJ - 2618 凸多边形 (半平面交)

题意：求n个凸多边形的交面积。

半平面交模板题。

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 typedef double db;
 const int N=+,inf=0x3f3f3f3f;
 const db pi=acos(-),eps=1e-;
 struct P {
     db x,y;
     P operator-(P b) {return {x-b.x,y-b.y};}
     P operator+(P b) {return {x+b.x,y+b.y};}
     P operator*(db b) {return {x*b,y*b};}
     bool operator<(const P& b)const {return x!=b.x?x<b.x:y<b.y;}
     bool operator==(const P& b)const {return x==b.x&&y==b.y;}
 } p[N],hpi[N];
 db cross(P a,P b) {return a.x*b.y-a.y*b.x;}
 struct Line {
     P p,v;
     db rad()const {return atan2(v.y,v.x);};
 } line[N],q[N];
 bool onleft(P p,Line a) {return cross(a.v,p-a.p)>;}
 P its(Line a,Line b) {
     P v=a.p-b.p;
     db t=cross(b.v,v)/cross(a.v,b.v);
     return a.p+a.v*t;
 }
 bool operator<(const Line& a,const Line& b) {
     db r1=a.rad(),r2=b.rad();
     return fabs(r1-r2)>eps?r1<r2:onleft(a.p,b);
 }
 int n,m,nl,nhpi;
 void HPI() {
     sort(line,line+nl);
     int hd=,tl=-;
     for(int i=; i<nl; ++i) {
         if(hd<=tl&&fabs(line[i].rad()-q[tl].rad())<eps)continue;
         if(hd<tl&&cross(q[hd].v,q[tl].v)<&&onleft(its(q[hd],q[tl]),line[i]))continue;
         for(; hd<tl&&!onleft(its(q[tl-],q[tl]),line[i]); --tl);
         for(; hd<tl&&!onleft(its(q[hd],q[hd+]),line[i]); ++hd);
         q[++tl]=line[i];
     }
     nhpi=;
     for(int i=hd; i<tl; ++i)hpi[nhpi++]=its(q[i],q[i+]);
     hpi[nhpi++]=its(q[tl],q[hd]);
 }
 db area() {
     db ret=;
     for(int i=; i<nhpi-; ++i)ret+=cross(hpi[i]-hpi[],hpi[i+]-hpi[]);
     return fabs(ret/);
 }

 int main() {
     for(scanf("%d",&m); m--;) {
         scanf("%d",&n);
         for(int i=; i<n; ++i)scanf("%lf%lf",&p[i].x,&p[i].y);
         for(int i=; i<n; ++i)line[nl++]= {p[i],p[(i+)%n]-p[i]};
     }
     HPI();
     printf("%.3f\n",area());
     return ;
 }