POJ2002 正方形的個數（二維點的哈希）

給定N個點，求出這些點一共可以構成多少個正方形。

若正方形爲ABCD，A座標爲(x1, y1)，B座標爲(x2, y2)，則很容易可以推出C和D的座標。對於特定的A和B座標，C和D可以在線段AB的左邊或者右邊，即有兩種情況。

因此只需要枚舉點A和點B，然後計算出兩種對應的C和D的座標，判斷是否存在即可。這樣計算完之後得到的答案是正確答案的4倍，因爲正方形的4條邊都枚舉了，所以答案要右移兩位。

#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;
}