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[];
 int n,t,ans,ed,tp,now,f[];double eps=1e-;char cmd[];

 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 ;
     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 ;
     return ;
 }

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