POJ——2002——Squares

Squares

Time Limit: 3500MS Memory Limit: 65536KTotal Submissions: 18697 Accepted: 7205

给出一些点，统计这些店中构成正方形的个数

利用向量已知对角线的两个顶点， 可求得构成正方形的另外两个顶点，可二分找点；

Description

A square is a 4-sided polygon whose sides have equal length and adjacent sides form 90-degree angles. It is also a polygon such that rotating about its centre by 90 degrees gives the same polygon. It is not the only polygon with the latter property, however, as a regular octagon also has this property.

So we all know what a square looks like, but can we find all possible squares that can be formed from a set of stars in a night sky? To make the problem easier, we will assume that the night sky is a 2-dimensional plane, and each star is specified by its x and y coordinates.

Input

The input consists of a number of test cases. Each test case starts with the integer n (1 <= n <= 1000) indicating the number of points to follow. Each of the next n lines specify the x and y coordinates (two integers) of each point. You may assume that the points are distinct and the magnitudes of the coordinates are less than 20000. The input is terminated when n = 0.

Output

For each test case, print on a line the number of squares one can form from the given stars.

Sample Input

41 00 11 10 090 01 02 00 21 22 20 11 12 14-2 53 70 05 20

Sample Output

161

Source

#include <iostream>#include <cstdio>#include <cstring>#include<algorithm>using namespace std;struct node{    int x,y;    bool operator <(const node &A)const    {        if(x==A.x)            return y<A.y;        return x<A.x;    }    bool operator != (const node &B)const    {        if(x!=B.x||y!=B.y)            return 1;        return 0;    }}que[10100];int main(){   int n;   int x,y;   int xmid,ymid;   node now,tmp;   while(cin>>n,n)   {       for(int i=0;i<n;i++)       {           scanf("%d%d",&que[i].x,&que[i].y);           que[i].x<<=1;que[i].y<<=1;       }       sort(que,que+n);       int ans=0;       for(int i=0;i<n;i++)       {           for(int j=i+1;j<n;j++)           {               xmid=(que[i].x+que[j].x)/2;               ymid=(que[i].y+que[j].y)/2;               x=(que[i].x-que[j].x)/2;               y=(que[i].y-que[j].y)/2;               now.x=xmid-y;               now.y=ymid+x;               tmp.x=(2*xmid)-now.x;               tmp.y=(2*ymid)-now.y;               int cnt=lower_bound(que,que+n,now)-que;               if(que[cnt]!=now)                continue;                cnt=lower_bound(que,que+n,tmp)-que;                if(que[cnt]!=tmp)                    continue;                ans++;           }       }       printf("%d\n",ans>>1);   }    return 0;}