poj 2002

给你一些点，求这些点中可以组成的正方形的个数

额，把点存起来，然后找的时候用Hash，依次枚举两个点（枚举任意两个点，其他两个点都可以确定，左边一个，右边一个）最后/4(每天边都枚举了4遍)

然后，就没有然后了

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;

const int N = 1010;
const int H = 10007;
int ptx[N], pty[N];

struct Node
{
	int x;
	int y;
	int next;
};
Node node[N];
int cur;
int n;
long ans;
int hashTable[H];

void initHash()
{
	for (int i = 0; i < H; ++i) hashTable[i] = -1;
	cur = 0;
	ans = 0;
}

void insertHash(int x, int y)
{
	int h = (x * x + y * y) % H;
	node[cur].x = x;
	node[cur].y = y;
	node[cur].next = hashTable[h];
	hashTable[h] = cur;
	++cur;
}

bool searchHash(int x, int y)
{
	int h = (x * x + y * y) % H;
	int next;
	next = hashTable[h];
	while (next != -1)
	{
		if (x == node[next].x && y == node[next].y) return true;
		next = node[next].next;
	}
	return false;
}

int main()
{
	while (scanf("%d", &n) != EOF && n)
	{
		initHash();
		for (int i = 0; i < n; ++i)
		{
			scanf("%d%d", &ptx[i], &pty[i]);
			insertHash(ptx[i], pty[i]);
		}
		for (int i = 0; i < n; ++i)
		{
			for (int j = i + 1; j < n; ++j)
			{
				int x1 = ptx[i] - (pty[i] - pty[j]);
				int y1 = pty[i] + (ptx[i] - ptx[j]);
				int x2 = ptx[j] - (pty[i] - pty[j]);
				int y2 = pty[j] + (ptx[i] - ptx[j]);
				if (searchHash(x1, y1) && searchHash(x2, y2)) ++ans;
			}
		}
		for (int i = 0; i < n; ++i)
		{
			for (int j = i + 1; j < n; ++j)
			{
				int x1 = ptx[i] + (pty[i] - pty[j]);
				int y1 = pty[i] - (ptx[i] - ptx[j]);
				int x2 = ptx[j] + (pty[i] - pty[j]);
				int y2 = pty[j] - (ptx[i] - ptx[j]);
				if (searchHash(x1, y1) && searchHash(x2, y2)) ++ans;
			}
		}
		ans >>= 2;
		printf("%ld\n", ans);
	}
	return 0;
}