POJ 2002 Squares

Description
A square is a 4-sided polygonwhose sides have equal length and adjacent sides form 90-degree angles. It isalso a polygon such that rotating about its centre by 90 degrees gives the samepolygon. It is not the only polygon with the latter property,
however, as aregular octagon also has this property.

So we all know what a square looks like, but can we find all possible squaresthat can be formed from a set of stars in a night sky? To make the problemeasier, we will assume that the night sky is a 2-dimensional plane, and eachstar is specified by its x and
y coordinates.
Input
The input consists of a numberof 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 thex and y coordinates (two integers) of each point. You may assume
that thepoints 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 aline 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

题目简介：给出n个点的坐标，判断这些点一共可以组成多少个正方形。
方法：哈希表。
正方形点的求法。已知（x1，y1），（x2，y2）。
则
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);

用(N.x*2+N.y+60000)%60000处理冲突。

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

class node
{
public:
int x, y;
}num[1010];

class HASH
{
public:
list<node> Hash[60000];

void Insert(node N)
{
Hash[(N.x*2+N.y+60000)%60000].push_back(N);
}

void Clear()
{
for(int i = 0;i<60000;i++)
{
Hash[i].clear();
}
}

int find(node N)
{
int target = (N.x*2+N.y+60000)%60000;
for(list<node>::iterator it = Hash[target].begin();it!=Hash[target].end();it++)
{
if(it->x==N.x&&it->y==N.y)
{
return 1;
}
}
return 0;
}
}My_hash;

int main()
{
int n, i, j;
while(scanf("%d",&n),n)
{
My_hash.Clear();
for(i = 0;i<n;i++)
{
node N;
scanf("%d%d",&N.x, &N.y);
num[i] = N;
My_hash.Insert(N);
}

int count = 0;
for(i = 0;i<n;i++)
{
for(j = i+1;j<n;j++)
{
node t1, t2;
t1.x=num[i].x + num[i].y - num[j].y;
t1.y=num[i].y - num[i].x + num[j].x;
t2.x=num[j].x + num[i].y - num[j].y;
t2.y=num[j].y - num[i].x + num[j].x;
if(My_hash.find(t1)&&My_hash.find(t2))
{
count++;
}

t1.x=num[i].x - num[i].y + num[j].y;
t1.y=num[i].y + num[i].x - num[j].x;
t2.x=num[j].x - num[i].y + num[j].y;
t2.y=num[j].y + num[i].x - num[j].x;
if(My_hash.find(t1)&&My_hash.find(t2))
{
count++;
}
}
}

printf("%d\n",count/4);
}
return 0;
}