poj 2002 Squares

亲测不hash会MLE= =

Squares

Time Limit: 3500MS		Memory Limit: 65536K
Total Submissions: 14856		Accepted: 5632

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

Source

Rocky Mountain 2004

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正方形(中文版)

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总时间限制:3500ms内存限制:65536kB描述

给定直角坐标系中的若干整点，请寻找可以由这些点组成的正方形，并统计它们的个数。

输入包括多组数据，每组数据的第一行是整点的个数n(1<=n<=1000)，其后n行每行由两个整数组成，表示一个点的x、y坐标。输入保证一组数据中不会出现相同的点，且坐标的绝对值小于等于20000。输入以一组n=0的数据结尾。输出对于每组输入数据，输出一个数，表示这组数据中的点可以组成的正方形的数量。样例输入

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

样例输出

1
6
1

我的解答

/*=============================================================================
#     FileName: square.cpp
#         Desc: poj 2002
#       Author: zhuting
#        Email: [email protected]
#     HomePage: my.oschina.net/locusxt
#      Version: 0.0.1
#    CreatTime: 2013-12-04 18:27:36
#   LastChange: 2013-12-04 19:35:39
#      History:
=============================================================================*/
#include <cstdio>
#include <cstdlib>
#include <string>
#include <cstring>
#include <algorithm>
#include <bitset>
#include <vector>
#define maxn 1005
#define maxl 40005
#define offset 20000/*偏移避免负数*/
using namespace std;

/*不hash会爆内存...*/

class point
{
	public:
		int x;
		int y;
		point()
		{
			x = 0;
			y = 0;
		}
};
point p[maxn];

class hash/*拉链法解决冲突的问题*/
{
	public:
		vector <int> vn;
};

hash h[maxl];



bool find(int x, int y)/*寻找是否有点(x,y)*/
{
	int len = h[x].vn.size();
	if (len == 0) return 0;
	for (int i = 0; i < len; ++i)
	{
		if (h[x].vn[i] == y)
			return 1;
	}
	return 0;
}

void add(int x, int y)/*加入点*/
{
	h[x].vn.push_back(y);
	return;
}

int main()
{
	int n = 0;
	int xtmp = 0, ytmp = 0;
	while(scanf("%d", &n) && n)
	{
		memset (p, 0, sizeof(p));
		memset (h, 0, sizeof(h));
		for (int i = 0; i < n; ++i)
		{
			scanf("%d%d", &xtmp, &ytmp);
			xtmp += offset;
			ytmp += offset;
			add(xtmp, ytmp);
			p[i].x = xtmp;
			p[i].y = ytmp;
		}
		/*没有必要排序*/
		//sort(p, p + n, cmpx);

		int square_num = 0;
		for (int i = 0; i < n; ++i)
			for (int j = i + 1; j < n; ++j)
			{
				int pix = p[i].x;
				int piy = p[i].y;
				int pjx = p[j].x;
				int pjy = p[j].y;
				int dy = pjy - piy;
				int dx = pjx - pix;
				/*分直线上下找*/
				if (find(pix - dy, piy + dx) && find(pjx - dy, pjy + dx))
				{
					++square_num;
				}
				if (find(pix + dy, piy - dx) && find(pjx + dy, pjy - dx))
				{
					++square_num;
				}
			}
		printf("%d\n", square_num / 4);/*存在重复*/
	}
	return 0;
}