poj 2002 Squares（枚举+点hash）

Squares

Time Limit: 3500MS Memory Limit: 65536KTotal Submissions: 12372 Accepted: 4535

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

Source

Rocky Mountain 2004

题目：http://poj.org/problem?id=2002

题意：给你n个点，问从n个点里面挑出4个点组成不同正方形的个数

分析：最容易想到的方法就是枚举两个点，那么另两个点就可以求出来，求出来后判断下这两个点是否存在，存在就+1，具体的判断只需要把所有点hash下就行

代码：

#include<cstdio>#include<iostream>#include<cstring>using namespace std;const int mm=1111;const int mod=100007;struct point{    int x,y;}g[mm];struct hashTable{    int h[mod],p[mod],size;    point s[mod];    int hash(int x,int y)    {        return ((x*131+y+mod)&0x7FFFFFFF)%mod;    }    void insert(int x,int y)    {        int i,id=hash(x,y);        for(i=h[id];i>=0;i=p[i])            if(s[i].x==x&&s[i].y==y)return;        s[size].x=x,s[size].y=y;        p[size]=h[id],h[id]=size++;    }    int find(int x,int y)    {        int i,id=hash(x,y);        for(i=h[id];i>0;i=p[i])            if(s[i].x==x&&s[i].y==y)return i;        return 0;    }    void clear()    {        size=1;        memset(h,-1,sizeof(h));    }}ht;bool check(point P,point Q){    int add1=P.y-Q.y,add2=Q.x-P.x;    return ht.find(P.x+add1,P.y+add2)&&ht.find(Q.x+add1,Q.y+add2);}int main(){    int i,j,n,ans;    while(scanf("%d",&n),n)    {        ht.clear();        for(i=0;i<n;++i)        {            scanf("%d%d",&g[i].x,&g[i].y);            ht.insert(g[i].x,g[i].y);        }        for(ans=i=0;i<n;++i)            for(j=0;j<n;++j)                if(i!=j&&check(g[i],g[j]))++ans;        printf("%d\n",ans/4);    }    return 0;}