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

1 #include
      
        2 #define FFC(i,a,b) for(int i=a;i<=b;++i) 3      4 struct line{
    double x1,y1,x2,y2;}a[1010]; 5 int n,t,ans,ed,tp,now,f[1010];double eps=1e-10;char cmd[2]; 6  7 int is(line a,line b){ 8     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; 9     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;10     return 1;11 }12 13 int find(int x){
    return f[x]==x?x:(f[x]=find(f[x]));}14 void merge(int x,int y){
    int xx,yy;if((xx=find(x))!=(yy=find(y)))f[xx]=yy;}15 16 int main(){17     scanf("%d",&t);18     while(t--){19         scanf("%d",&n),ed=0;20         FFC(i,1,n)f[i]=i;21         FFC(i,1,n){22             scanf("%s",cmd);23             if(cmd[0]=='P'){24                 ed++,scanf("%lf%lf%lf%lf",&a[ed].x1,&a[ed].y1,&a[ed].x2,&a[ed].y2);25                 FFC(j,1,ed-1)if(is(a[j],a[ed]))merge(ed,j);26             }else {27                 scanf("%d",&tp),ans=0,now=find(tp);28                 FFC(i,1,ed)if(find(i)==now)ans++;29                 printf("%d\n",ans);30             }31         }32         if(t)puts("");33     }34     return 0;35 }