poj-2002 Squares

Squares

Time Limit: 3500MS		Memory Limit: 65536K
Total Submissions: 11102		Accepted: 4021

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<cmath>
int n,i,j,ans;
using namespace std;
struct set
{
	int x,y;
};
set point[1005];
set c,d;
int cmp(const void *a,const void *b)
{
	if(((set *)a)->x==((set *)b)->x)
		return ((set *)a)->y-((set *)b)->y;
	return ((set *)a)->x-((set *)b)->x;
}
int fab(int a)
{
	return a>0?a:-a;
}
bool search(set p,int n)
{
	int max=n,min=1,mid;
	while(max>=min)
	{
		mid=(max+min)/2;
		if(cmp(&point[mid],&p)>0)
		{
			max=mid-1;
		}
		else if(cmp(&point[mid],&p)<0)
		{
			min=mid+1;
		}
		else
			return true;
	}
	return false;
}
int main()
{
	while(~scanf("%d",&n))
	{
		ans=0;
		if(n==0)
			break;
		for(i=1;i<=n;i++)
			scanf("%d%d",&point[i].x,&point[i].y);
		qsort(point+1,n,sizeof(set),cmp);
		for(i=1;i<=n;i++)
		{
			for(j=i+1;j<=n;j++)
			{
				int sign=(point[i].x-point[j].x)*(point[i].y-point[j].y);
				c.x=point[i].x+fab(point[j].y-point[i].y);
				c.y=point[i].y+fab(point[j].x-point[i].x);
				d.x=point[j].x+fab(point[j].y-point[i].y);
				d.y=point[j].y+fab(point[j].x-point[i].x);
				if(sign<0)
				{
					if(search(c,n) && search(d,n))
						ans++;
				}
				else if(sign==0 && point[i].x==point[j].x)
				{
					if(search(c,n) && search(d,n))
						ans++;
				}
			}
		}
		printf("%d\n",ans);
	}
	return 0;
}