题目大意：给定n个凸多边形，求交集的面积

时隔多年我终于把完整的半平面交搞出来了……真尼玛艰辛……

曾经写了一发 RE到死 于是就搁置0.0 今天写一发又是WA到死的节奏……

不多说直接上代码 其实刘汝佳同学写麻烦了 每次插入一个半平面之后不用两端都删的 只删一端 最后再处理两端的部分就行

300题留念……切了道模板题也不错

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define M 510
#define EPS 1e-7
using namespace std;
struct point{
	double x,y;
	point(){}
	point(double _,double __):
		x(_),y(__){}
	point operator + (const point &p) const
	{
		return point(x+p.x,y+p.y);
	}
	point operator - (const point &p) const
	{
		return point(x-p.x,y-p.y);
	}
	double operator * (const point &p) const
	{
		return x*p.y-y*p.x;
	}
	point operator * (const double &rate) const
	{
		return point(x*rate,y*rate);
	}
	void Read()
	{
		scanf("%lf%lf",&x,&y);
	}
}points[M];
struct line{
	point p,v;
	double alpha;
	line(){}
	line(const point &p1,const point &p2):p(p1),v(p2-p1)
	{
		alpha=atan2(v.y,v.x);
	}
	bool operator < (const line &l) const
	{
		return alpha < l.alpha;
	}
}lines[M];
int n,m,tot;
line *q[M];
int r,h;
double ans;
bool Onleft(const point &p,const line &l)
{
	return (l.p-p)*l.v>=0;
}
point Get_Intersection(const line &l1,const line &l2)
{
	point u=l1.p-l2.p;
	double temp=(l2.v*u)/(l1.v*l2.v);
	return l1.p+l1.v*temp;
}
void Get_Half_Plane_Intersection()
{
	int i;
	for(i=1;i<=tot;i++)
	{
		while( r-h>=2 && !Onleft(Get_Intersection(*q[r],*q[r-1]),lines[i]) )
			q[r--]=0x0;
		if( r-h>=1 && fabs(lines[i].v*q[r]->v)<=0 )
			q[r]=Onleft(lines[i].p,*q[r])?&lines[i]:q[r];
		else q[++r]=&lines[i];
	}
	while( r-h>=2 && !Onleft(Get_Intersection(*q[h+1],*q[h+2]),*q[r]) )
		q[++h]=0x0;
	while( r-h>=2 && !Onleft(Get_Intersection(*q[r],*q[r-1]),*q[h+1]) )
		q[r--]=0x0;
}
int main()
{
	int i,j;
	cin>>n;
	for(i=1;i<=n;i++)
	{
		point first,p1,p2;
		scanf("%d",&m);
		first.Read();p2=first;
		for(j=2;j<=m;j++)
		{
			p1=p2;p2.Read();
			lines[++tot]=line(p1,p2);
		}
		lines[++tot]=line(p2,first);
	}
	sort(lines+1,lines+tot+1);
	Get_Half_Plane_Intersection();
	if(r-h<=2)
		return puts("0.000"),0;
	tot=0;
	for(i=h+2;i<=r;i++)
		points[++tot]=Get_Intersection(*q[i],*q[i-1]);
	points[++tot]=Get_Intersection(*q[r],*q[h+1]);
	for(i=2;i<=tot;i++)
		ans+=points[i-1]*points[i];
	ans+=points[tot]*points[1];
	printf("%.3lf\n",ans/2);
	return 0;
}

BZOJ 2618 CQOI2006 凸多边形 半平面交