Squares<哈希>

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


就是看看给出的点能组成几个正方形；
不要四个四个的组合，想也会超时，任意两个组合，然后计算出另外两个点；
再去查询是否存在；结果要除以4的；

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
bool x[5000][5000];//我才不告诉你其实数据是-2000--2000
int si[1010],sj[1010];
int f(int a,int b)
{
    if(x[a+2500][b+2500])
        return 1;
    return 0;
}
int main()
{
    int a,b,i,j,n,sum;
    int x1,y1,x2,y2,x3,y3,x4,y4;
    while(scanf("%d",&n)&&n)
    {
        sum=0;
        memset(x,0,sizeof(x));
        for(i=0; i<n; i++)
        {
            scanf("%d %d",&a,&b);
            si[i]=a;
            sj[i]=b;
            x[a+2500][b+2500]=1;
        }
        for(i=1; i<n; i++)
        {
            x1=si[i];
            y1=sj[i];
            for(j=0; j<i; j++)
            {
                x2=si[j];
                y2=sj[j];
                x3=x1+(y1-y2);//计算剩下两个点
                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);//记得除以4
    }
    return 0;
}