Poj 2002 && 3432

来源：互联网发布：在爱奇艺网络剧工作编辑：程序博客网时间：2024/05/16 19:30

1Y*2

相同题目

排序点

二分查找

n^2枚举对角线

#include<vector>#include<algorithm>#include<stdio.h>struct point{int x,y;bool operator <(const point &tp)const{if(x!=tp.x)return x<tp.x;return y<tp.y;}bool operator ==(const point &tp)const{return tp.x==x&&tp.y==y;}}tp;std::vector<point>L;int bfind(point t){int l=0,r=L.size()-1;while(r>l){if(r-l==1){if(t==L[r]||t==L[l])return 1;return 0;}int mid=(l+r)/2;if(L[mid]==t)return 1;else if(L[mid]<t)l=mid;else r=mid;}}int main(){int n;while(scanf("%d",&n)!=EOF&&n){int i,j;L.clear();for(i=0;i<n;i++){scanf("%d%d",&tp.x,&tp.y);tp.x*=2;tp.y*=2;L.push_back(tp);}std::sort(L.begin(),L.end());int res=0;for(i=0;i<L.size();i++)for(j=i+1;j<L.size();j++){int x1=L[i].x,x2=L[j].x,y1=L[i].y,y2=L[j].y;tp.x=(x1+x2+y1-y2)/2;tp.y=(y1+y2+x2-x1)/2;if(bfind(tp)){tp.x=(x1+x2-y1+y2)/2;tp.y=(y1+y2-x2+x1)/2;if(bfind(tp)){//printf("%d %d\n",tp.x,tp.y);res++;}}}res/=2;printf("%d\n",res);}return 0;}