hash poj2002

题意：问n个点中可以组成多少个正方形，n<1000

思路：一开始TLE了，是hash的边，后来看网上是hash的点，hash函数是（x*x+y*y）%prime

已知正方形的两个点，求另外两个坐标：(x1,y1) 、(x2,y2)

则： x3=x1+(y1-y2)   y3=y1-(x1-x2)

       x4=x2+(y1-y2)   y4=y2-(x1-x2)

#include<stdio.h>
#include<vector>
#include<string.h>
#include<iostream>
using namespace std;

const int N=1005;
const int prime=2003;

int x[N],y[N];
vector<int> hash[prime];

bool search(int _x,int _y){
    int key=(_x*_x+_y*_y)%prime;
    int len=hash[key].size();
    for(int i=0;i<len;i++){
        int v=hash[key][i];
        if(x[v]==_x&&y[v]==_y) return 1;
    }
    return 0;
}

int main(){
    int i,j,k;
    int n;
    while(scanf("%d",&n)!=EOF&&n){
        for(i=0;i<prime;i++) hash[i].clear();
        for(i=0;i<n;i++){
            scanf("%d%d",&x[i],&y[i]);
            int key=(x[i]*x[i]+y[i]*y[i])%prime;
            hash[key].push_back(i);
        }
        int ans=0;
        int x1,y1,x2,y2;
        for(i=0;i<n;i++)
            for(j=i+1;j<n;j++){
                if (i == j) continue;
                x1=x[i]+y[i]-y[j];
                y1=y[i]-x[i]+x[j];
                x2=x[j]+y[i]-y[j];
                y2=y[j]-x[i]+x[j];
                if (search(x1, y1) && search(x2, y2)) ans++;
                x1=x[i]-y[i]+y[j];
                y1=y[i]+x[i]-x[j];
                x2=x[j]-y[i]+y[j];
                y2=y[j]+x[i]-x[j];
                if (search(x1, y1) && search(x2, y2)) ans++;
            }
        printf("%d\n",ans/4);
    }
    return 0;
}