POJ2002 Squares

Squares

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

给出n个整点，求正方形的个数n的范围是1000，所以n*n左右的算法于是乎，枚举两个点，检测另外两个点是否在检测呢就用hash

#include <iostream>
#include <cstdio>
#include <cstring>
#define MOD 4209207
#define N 1010
#define M 20000
using namespace std;

struct CHash
{
int m_nX;
int m_nY;
bool m_bExi;
} nhash[MOD];

inline int hash(int x,int y)
{
int k = x * M + y;
int p = (k<0?-k:k) % MOD;
while( nhash[ p ].m_bExi && ( nhash[ p ].m_nX != x || nhash[ p ].m_nY != y ) )
{
p ++;
if( p >= MOD ) p %= MOD;
}
return p;
}
int x
,y
;
int main()
{
int n,tx,ty,dx,dy,cx,cy;
while(scanf("%d",&n)==1&&n)
{
int ans = 0;
memset(nhash,0,sizeof(nhash));
for(int i = 0 ; i < n ; i++ )
{
scanf("%d %d",x+i,y+i);
int k = hash(x[i],y[i]);
nhash[ k ].m_nX = x[i];
nhash[ k ].m_nY = y[i];
nhash[ k ].m_bExi = true;
}
for(int i = 0 ; i < n ; i++ )
{
//这里呢，另有一番说法...以下摘自poj discuss里
/*******************
我开始用的公式是
x3=x1+(y1-y2);y3=y1-(x1-x2); x4=x2+(y1-y2);y4=y2-(x1-x2)
和你同样的处理方法（j从i+1到n循环），最后就WA

经我仔细研究发现：
其实对于这个公式求的正方形只能是P1P2P4P3形状的
这样如果用第一种处理方式，就很可能处理掉很多正方形，其实任何给定的P1P2，本应该能为某个正方形贡献一次重复，那么四条边的话就重复四次，最后就除以4
但是我们输入的P1P2的顺序可能使我们得不到一次重复 我们误以为顺序和反过来一样 就对半分 除以2 其实是错的
如果我们把题意总第一组测试实例4个点的顺序改变一下再输入，先不要除 你会发现得到的的正方形的个数是不固定的 这就是由于输入的顺序照成的

那么第二种处理方式（就是j从0到n循环）正好弥补了这个错误
*********************************/
for(int j = 0 ; j < n ; j++ )
{
if(i==j)continue;
tx = x[ j ] - x[ i ];
ty = y[ j ] - y[ i ];
cx = x[ i ] + ty;
cy = y[ i ] - tx;
dx = x[ j ] + ty;
dy = y[ j ] - tx;
if( nhash[ hash(cx,cy) ].m_bExi && nhash[ hash(dx,dy) ].m_bExi )
{
ans++;
}
}
}
printf("%d\n",ans>>2);
}
return 0;
}