Squares（枚举+set 查找）

http://poj.org/problem?id=2002

题意：给出n组坐标，判断这些坐标能组成的正方形的个数。

思路：参考某大神的想法，先枚举两个点，然后利用公式表示出另外两个点，判断这两个点是否在这n组坐标中,其中查找另两个坐标用的set容器。

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);

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <set>
const int N=100000;
const int maxn=1010;

using namespace std;
struct node
{
    long long  x;
    long long  y;
} p[maxn];
int main()
{
    int n;
    while(~scanf("%d",&n)&&n)
    {
        int ans = 0;
        set<long long>tt;
        for (int i = 0; i < n; i ++)
        {
            scanf("%lld %lld",&p[i].x,&p[i].y);
            tt.insert(p[i].x*N+p[i].y);//将坐标转化成一个数，坐标不同，则这个数就不同
        }
        for (int i = 0; i < n; i ++)
        {
            int x1 = p[i].x;
            int y1 = p[i].y;
            for (int j = i+1; j < n; j ++)
            {
                int x2 = p[j].x;
                int y2 = p[j].y;
                int x3 = x1+(y1-y2);
                int y3 = y1-(x1-x2);
                int x4 = x2+(y1-y2);
                int y4 = y2-(x1-x2);
                if (tt.count(x3*N+y3) && (tt.count(x4*N+y4)))// 查找坐标是否存在
                    ++ans;
                x3 = x1-(y1-y2);
                y3 = y1+(x1-x2);
                x4 = x2-(y1-y2);
                y4 = y2+(x1-x2);
                if (tt.count(x3*N+y3) && (tt.count(x4*N+y4)))
                    ++ans;
            }
        }
        printf("%d
",ans/4);
    }
    return 0;
}