hdu_1558_Segment set(並查集+計算幾何)
阿新 • • 發佈:2019-02-19
題目連線:http://acm.hdu.edu.cn/showproblem.php?pid=1558
題意:P為畫線段,Q為詢問當前這條線段所在的集合有多少線段
題解:如果兩條線段有交點,那麼就連線這兩個集合
#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\n",ans); } } if(t)puts(""); } return 0; }