BZOJ

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=1000+10,inf=0x3f3f3f3f;
const db pi=acos(-1),eps=1e-8;
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)>0;}
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=0,tl=-1;
    for(int i=0; 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)<0&&onleft(its(q[hd],q[tl]),line[i]))continue;
        for(; hd<tl&&!onleft(its(q[tl-1],q[tl]),line[i]); --tl);
        for(; hd<tl&&!onleft(its(q[hd],q[hd+1]),line[i]); ++hd);
        q[++tl]=line[i];
    }
    nhpi=0;
    for(int i=hd; i<tl; ++i)hpi[nhpi++]=its(q[i],q[i+1]);
    hpi[nhpi++]=its(q[tl],q[hd]);
}
db area() {
    db ret=0;
    for(int i=1; i<nhpi-1; ++i)ret+=cross(hpi[i]-hpi[0],hpi[i+1]-hpi[0]);
    return fabs(ret/2);
}

int main() {
    for(scanf("%d",&m); m--;) {
        scanf("%d",&n);
        for(int i=0; i<n; ++i)scanf("%lf%lf",&p[i].x,&p[i].y);
        for(int i=0; i<n; ++i)line[nl++]= {p[i],p[(i+1)%n]-p[i]};
    }
    HPI();
    printf("%.3f
",area());
    return 0;
}