hdu_1558_Segment set(并查集+计算几何)

题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=1558

题意：P为画线段，Q为询问当前这条线段所在的集合有多少线段

题解：如果两条线段有交点，那么就连接这两个集合

#include<cstdio>
#define FFC(i,a,b) for(int i=a;i<=b;++i)
    
struct line{double x1,y1,x2,y2;}a[1010];
int n,t,ans,ed,tp,now,f[1010];double eps=1e-10;char cmd[2];

int is(line a,line b){
    if(((a.x2-a.x1)*(b.y1-a.y1)-(b.x1-a.x1)*(a.y2-a.y1))*((a.x2-a.x1)*(b.y2-a.y1)-(b.x2-a.x1)*(a.y2-a.y1))>eps)return 0;
    if(((b.x2-b.x1)*(a.y1-b.y1)-(a.x1-b.x1)*(b.y2-b.y1))*((b.x2-b.x1)*(a.y2-b.y1)-(a.x2-b.x1)*(b.y2-b.y1))>eps)return 0;
    return 1;
}

int find(int x){return f[x]==x?x:(f[x]=find(f[x]));}
void merge(int x,int y){int xx,yy;if((xx=find(x))!=(yy=find(y)))f[xx]=yy;}

int main(){
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n),ed=0;
        FFC(i,1,n)f[i]=i;
        FFC(i,1,n){
            scanf("%s",cmd);
            if(cmd[0]=='P'){
                ed++,scanf("%lf%lf%lf%lf",&a[ed].x1,&a[ed].y1,&a[ed].x2,&a[ed].y2);
                FFC(j,1,ed-1)if(is(a[j],a[ed]))merge(ed,j);
            }else {
                scanf("%d",&tp),ans=0,now=find(tp);
                FFC(i,1,ed)if(find(i)==now)ans++;
                printf("%d
",ans);
            }
        }
        if(t)puts("");
    }
    return 0;
}