bzoj 2618 半平面交

觉得巨恶心不知道怎么办，结果一看题解发现因为凸多边形相交还是一个多边形，并不会变成多个奇怪的很多个小图形，所以可以直接半平面交求面积，测板子+1，过样例就A真爽。

#include<bits/stdc++.h>
#define maxl 1010
#define eps 1e-8
using namespace std;

inline int sgn(double x)
{
	if(x>-eps && x<eps) return 0;
	if(x>0) return 1;
	else	return -1;
}

struct point
{
	double x,y;
	point(double a=0,double b=0)
	{
		x=a;y=b;
	}
	point operator + (const point &b)const
	{
		return point(x+b.x,y+b.y);
	}
	point operator - (const point &b)const
	{
		return point(x-b.x,y-b.y);
	}
	friend point operator * (double t,point p)
	{
		return point(t*p.x,t*p.y);
	}
};
inline double dot(const point &a,const point &b)
{
	return a.x*b.x+a.y*b.y;
}
inline double det(const point &a,const point &b)
{
	return a.x*b.y-a.y*b.x;
}
struct halfplane
{
	point s,e;
	double k;
	halfplane(point a=point(),point b=point())
	{
		s=a;e=b;
		k=atan2(e.y-s.y,e.x-s.x);
	}
	point operator & (const halfplane &b)const
	{
		double t=det(b.s-s,b.e-b.s);
		t=t/det(e-s,b.e-b.s);
		return s+t*(e-s);
	}
};

int n,m[maxl],cnt;
point a[maxl][maxl],res[maxl];
halfplane line[maxl*maxl];
double ans;

inline void prework()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&m[i]);
		for(int j=1;j<=m[i];j++)
			scanf("%lf%lf",&a[i][j].x,&a[i][j].y);
		a[i][0]=a[i][m[i]];
		for(int j=0;j<m[i];j++)
			line[cnt++]=halfplane(a[i][j],a[i][j+1]);
	}
}

inline bool satisfy(point a,const halfplane L)
{
	return sgn(det(a-L.s,L.e-L.s))<=0;
}
inline bool HPIcmp(const halfplane &a,const halfplane &b)
{
	int res=sgn(a.k-b.k);
	return res==0 ? det(a.s-b.s,b.e-b.s)<0 : a.k<b.k;
}
halfplane Q[maxl];
inline void HPI(halfplane line[],int n,point res[],int &resn)
{
	int tot=n;
	sort(line,line+n,HPIcmp);
	tot=1;
	for(int i=1;i<n;i++)
	if(sgn(line[i].k-line[i-1].k)!=0)
		line[tot++]=line[i];
	int head=0,tail=1;
	Q[0]=line[0];Q[1]=line[1];resn=0;
	for(int i=2;i<tot;i++)
	{
		if(sgn(det(Q[tail].e-Q[tail].s,Q[tail-1].e-Q[tail-1].s))==0
		|| sgn(det(Q[head].e-Q[head].s,Q[head+1].e-Q[head+1].s))==0)
			return;
		while(head<tail && !satisfy(Q[tail]&Q[tail-1],line[i]))
			tail--;
		while(head<tail && !satisfy(Q[head]&Q[head+1],line[i]))
			head++;
		Q[++tail]=line[i];
	}
	while(head<tail && !satisfy(Q[tail]&Q[tail-1],Q[head]))
		tail--;
	while(head<tail && !satisfy(Q[head]&Q[head+1],Q[tail]))
		head++;
	if(tail<=head+1) return;
	for(int i=head;i<tail;i++)
		res[resn++]=Q[i]&Q[i+1];
	if(head<tail-1);
		res[resn++]=Q[head]&Q[tail];
}

inline double get_area(point a[],int n)
{
	double area=0;
	a[n]=a[0];
	for(int i=0;i<n;i++)
		area+=det(a[i],a[i+1]);
	return fabs(area/2);
}

inline void mainwork()
{
	int resn=0;
	HPI(line,cnt,res,resn);
	ans=get_area(res,resn);
}

inline void print()
{
	printf("%.3f",ans);
}

int main()
{
	prework();
	mainwork();
	print();
	return 0;
}