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[][];//我才不告诉你其实数据是-2000--2000
 int si[],sj[];
 int f(int a,int b)
 {
     if(x[a+][b+])
         return ;
     return ;
 }
 int main()
 {
     int a,b,i,j,n,sum;
     int x1,y1,x2,y2,x3,y3,x4,y4;
     while(scanf("%d",&n)&&n)
     {
         sum=;
         memset(x,,sizeof(x));
         for(i=; i<n; i++)
         {
             scanf("%d %d",&a,&b);
             si[i]=a;
             sj[i]=b;
             x[a+][b+]=;
         }
         for(i=; i<n; i++)
         {
             x1=si[i];
             y1=sj[i];
             for(j=; 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\n",sum/);//记得除以4
     }
     return ;
 }