这个题简直丧心病狂啊!!只要两个线段挨上就是一堆里的!!挨上,你懂么你懂么?相交算挨上,共线有重合算挨上,共线没重合不算!!

最讨厌数学了!讨厌讨厌@。@

还有 之前在纠结到底是以点作为单元还是线段作为单元呢?答案是线段套着点!

问一堆里的有多少怎么办?发现就加上==

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int pre[1010],sum[1010];
struct point{
	double x,y;
};
struct EDGE{
	point a,b;
} edge[1010];
int E;//边数
int Find(int x){
	return x==pre[x]? x:pre[x]=Find(pre[x]);
}
void Merge(int a,int b){
	int x=Find(a),y=Find(b);
	if(x!=y){
		pre[y]=x;
		sum[x]+=sum[y];
	}
}
double xmult(point a,point b,point c){//大于零代表a,b,c左转
	return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x);
}
bool OnSegment(point a,point b,point c){		//a,b,c共线时有效
	return c.x>=min(a.x,b.x)&&c.x<=max(a.x,b.x)&&c.y>=min(a.y,b.y)&&c.y<=max(a.y,b.y);
}
bool Cross(point a,point b,point c,point d){//判断ab 与cd是否相交
	double d1,d2,d3,d4;
	d1=xmult(c,d,a);
	d2=xmult(c,d,b);
	d3=xmult(a,b,c);
	d4=xmult(a,b,d);
	if(d1*d2<0&&d3*d4<0)	return 1;
	else	if(d1==0&&OnSegment(c,d,a))	return 1;
	else	if(d2==0&&OnSegment(c,d,b))	return 1;
	else	if(d3==0&&OnSegment(a,b,c))	return 1;
	else	if(d4==0&&OnSegment(a,b,d))	return 1;
	return 0;
}
int main()
{
	int i,j,k,T,n;
	char s[10];
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d",&n);E=0;
		for(i=1;i<=n;i++)	pre[i]=i,sum[i]=1;
		for(i=1;i<=n;i++)
		{
			scanf("%s",s);
			if(s[0]=='P'){
				E++;
				scanf("%lf%lf%lf%lf",&edge[E].a.x,&edge[E].a.y,&edge[E].b.x,&edge[E].b.y);
				for(j=1;j<E;j++)
					if(Find(E)!=Find(j)&&Cross(edge[E].a,edge[E].b,edge[j].a,edge[j].b))	Merge(E,j);
			}
			else	if(s[0]=='Q'){
				scanf("%d",&k);
				printf("%d\n",sum[Find(k)]);
			}
		}
		if(T)	printf("\n");
	}
	return 0;
}