Squares 四边形数量，求坐标很重要

Problem 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

4 1 0 0 1 1 1 0 0 9 0 0 1 0 2 0 0 2 1 2 2 2 0 1 1 1 2 1 4 -2 5 3 7 0 0 5 2 0

 

Sample Output

1 6 1

***************************************************************************************************************************

记着标记数组要用boolean类型

***************************************************************************************************************************

#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<queue>
#include<cstdio>
using namespace std;
int x[1010],y[1010];
bool fs[5000][5000];//此处要定义成bool类型的，否则超内存
int a,b,c,i,j,sum,n;
int f(int a,int b)
{
      if(fs[a+2500][b+2500])
         return 1;
     return 0;
}

int main()
{
    int x1,y1,x2,y2,x3,y3,x4,y4;
    int a,b,i,j,sum,n;
    while(scanf("%d",&n)&&n)
    {
        sum=0;
        memset(fs,0,sizeof(fs));
        for(i=0;i<n;i++)
        {
            scanf("%d %d",&a,&b);
            x[i]=a;
            y[i]=b;
            fs[a+2500][b+2500]=1;
        }
        for(i=0;i<n;i++)
        {
             x1=x[i];
             y1=y[i];
            for(j=0;j<i;j++)
            {
                 x2=x[j];
                 y2=y[j];
                 x3,y3,x4,y4;
                x3=x1+(y1-y2);//这个公式推推就出来啦
                y3=y1-(x1-x2);
                x4=x2+(y1-y2);
                y4=y2-(x1-x2);
                if(f(x3,y3)&&f(x4,y4))
                  sum++;
                x3=x1-(y1-y2);
                y3=y1+(x1-x2);
                x4=x2-(y1-y2);
                y4=y2+(x1-x2);
                if(f(x3,y3)&&f(x4,y4))
                  sum++;
            }
        }
        printf("%d
",sum/4);//注意此处要除以4，

    }
}