poj Squares 2002 （STL&&枚举+转换） 统计正方形的个数

Squares

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

//题意：

给你n个点，让你求出这n个点能组成几个正方形。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define N 1010
using namespace std;
struct zz
{
	int x;
	int y;
	friend bool operator<(zz a,zz b)
	{
		return a.x<b.x||(a.x==b.x&&a.y<b.y);
	}
}q[N<<1],tmp;
int main()
{
	int n,i,j,k;
	while(scanf("%d",&n),n)
	{
		for(i=0;i<n;i++)
			scanf("%d%d",&q[i].x,&q[i].y);
		sort(q,q+n);
		int cnt=0;
		for(i=0;i<n;i++)
		{
			for(j=i+1;j<n;j++)
			{
				tmp.x=q[i].x+q[i].y-q[j].y;
				tmp.y=q[i].y+q[j].x-q[i].x;
				if(!binary_search(q,q+n,tmp))
					continue;
				tmp.x=q[j].x+q[i].y-q[j].y;
				tmp.y=q[j].y+q[j].x-q[i].x;
				if(!binary_search(q,q+n,tmp))
					continue;
				cnt++;
			}
		}
		printf("%d\n",cnt/2);
	}
	return 0;
}