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

41 00 11 10 090 01 02 00 21 22 20 11 12 14-2 53 70 05 20

Sample Output

161

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 20000using 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[N],y[N];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;}