POJ_2002Squares解题报告

Squares

Time Limit: 3500MS		Memory Limit: 65536K
Total Submissions: 11181		Accepted: 4064

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

题目连接：http://poj.org/problem?id=2002

算法类型：Hash

解题思路：先把每个点所对应的Hash值储存在对应的容器里，在依次枚举两个点，算出其所对应的另外四个点，分别在其对应的容器中依次查找，如果两个点都找到，总数加一，最后把总数除以四。

算法实现：

#include
#include
#include
#include
#include
using namespace std;
#define MAX 9997
struct Squares
{
	int x;
	int y;
}squ[100002];
int Hash(Squares squ) //hash 函数；
{
	int s;
	s=squ.x*squ.x+squ.y*squ.y+squ.y;
	return s%MAX;
}
int  main()
{
	int N;
	while(scanf("%d",&N)!=EOF)
	{
		if(N==0)
		break;
		memset(squ,0,100002);
		vector v[MAX];  //系统容器；
		memset(v,0,MAX);
		int sum=0;
		int i,J,JJ,P,PP;
		int hash,hash1,hash2,hash3,hash4;
		int X1,Y1,X2,Y2,X3,Y3,X4,Y4;
		for(i=0;i