poj 2002 Squares

Squares

Time Limit: 3500MS		Memory Limit: 65536K
Total Submissions: 10585		Accepted: 3825

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

#include<iostream>
#include<algorithm>
using namespace std;
struct node
{
    int x;
    int y;
}star[1001];

int comp(node a,node b)
{
    if(a.x==b.x)
        return a.y<b.y;
    return a.x<b.x;
}

int main()
{
    int n;
    int i,j;
    int k;
    int sum;
//    freopen("E:\\c++\\oj\\t.txt","rt",stdin);
    while(1)
    {
        cin>>n;
        if(n==0)
            break;
        for(i=0;i<n;i++)
        {
            cin>>star[i].x>>star[i].y;
        }
        k=0;
        sum=0;
        sort(star,star+n,comp);
        for(i=0;i<n-1;i++)
        {
            for(j=i+1;j<n;j++)
            {
                node c,d;
                c.x=star[i].x+star[j].y-star[i].y;
                c.y=star[i].y+star[i].x-star[j].x;
                d.x=star[j].x+star[j].y-star[i].y;
                d.y=star[j].y+star[i].x-star[j].x;
                int res=binary_search(star,star+n,c,comp);
                if(res==0)
                    continue;
                int res2=binary_search(star,star+n,d,comp);
                if(res2==0)
                    continue;
                if(res && res2)
                    sum++;
            }
        }
        cout<<sum/2<<endl;
    }
    return 0;
}