Squares

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本文介绍了一种通过哈希表快速查找平面上由星星坐标形成的正方形的方法。利用点坐标计算并验证潜在的正方形顶点，有效解决了大规模点集中的几何形状识别问题。

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Squares

Time Limit: 3500MS		Memory Limit: 65536K
Total Submissions: 16784		Accepted: 6389

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

Sample Output

1
6
1

题意：在所给的点中找能组成的正方形的数目，

将给出的点都存入哈希表中，因为若是枚举4个点肯定会超时，可以枚举两个点，再以此求出能跟这两个点组成正方形的另外两个点，判断哈希表中是否存在这两个点

依据公式：

x3=x1+(y1-y2) ; y3=y1-(x1-x2) ;

x4=x2+(y1-y2) ; y4=y2-(x1-x2) ;

或

x3=x1-(y1-y2) ; y3=y1+(x1-x2) ;

x4=x2-(y1-y2) ; y4=y2+(x1-x2) ;

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <algorithm>
#include <string.h>

using namespace std;

const int prime=1999;

struct node
{
	int x;
	int y;
}point[1010];

typedef struct HASH
{
	int x;
	int y;
	HASH *next;
	HASH()
	{
		next=0;
	}
}Hash;

Hash *Hs[prime];//定义指针数组，作为哈希链表的头指针

void Insert(int k)//插入哈希表，标记散点
{
	int key=(point[k].x*point[k].x+point[k].y*point[k].y)%prime+1;
	if(!Hs[key])//若为空，直接将关键字插入
	{
		Hash *t=new Hash;
		t->x=point[k].x;
		t->y=point[k].y;
		Hs[key]=t;
	}
	else//若不为空，找到链表的最后节点再插入
	{
		Hash *t=Hs[key];
		while(t->next)
		{
			t=t->next;
		}
		t->next=new HASH;
		t->next->x=point[k].x;
		t->next->y=point[k].y;
	}
}

int Find(int x,int y)//查找该点是否在哈希表中
{
	int key=(x*x+y*y)%prime+1;
	if(!Hs[key])//头节点为空直接返回0
	{
		return 0;
	}
	else
	{
		Hash *t=Hs[key];
		while(t)//依次遍历
		{
			if(t->x==x&&t->y==y)
				return 1;
			t=t->next;
		}
	}
	return 0;
}

int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		int sum=0;
		if(!n)
			break;
		memset(Hs,0,sizeof(Hs));
		for(int i=1;i<=n;i++)
		{
			scanf("%d%d",&point[i].x,&point[i].y);
			Insert(i);//插入哈希表
		}
		for(int i=1;i<n;i++)
		{
			for(int j=i+1;j<=n;j++)
			{
				int s=point[i].x-point[j].x;
				int t=point[i].y-point[j].y;
				int x3=point[i].x+t;
				int y3=point[i].y-s;
				int x4=point[j].x+t;
				int y4=point[j].y-s;
				if(Find(x3,y3)&&Find(x4,y4))//判断另外两个点是否在哈希表中
					sum++;
				x3=point[i].x-t;
				y3=point[i].y+s;
				x4=point[j].x-t;
				y4=point[j].y+s;
				if(Find(x3,y3)&&Find(x4,y4))
					sum++;
			}
		}
		printf("%d\n",sum/4);
	}
	return 0;
}