（POJ - 2002）Squares

（POJ - 2002）Squares

Time Limit: 3500MS Memory Limit: 65536K

Total Submissions: 20599 Accepted: 7919

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

题目大意：给出n个点的坐标，问能组成多少个正方形。

思路：直接暴力枚举。先任意选定两个点(x1,y1),(x2,y2)，假设以这两个点为边能组成正方形，那么其余两个点可以通过全等三角形的性质计算出，画个图也很容易出来，有两种情况（上面一个，下面一个）。 只要判断理论计算出来的两点是否存在，存在
则ans++，由于这里枚举的时候正方形的每一条边长都枚举了一边所以最后ans/4才是答案。
细节见代码。
ps：尤其注意这里用memset会wa。

#include<cstdio>
#include<cstring>
using namespace std;

const int maxn=20005;
bool vis[maxn<<1][maxn<<1];

struct node
{
    int x,y;
}a[1005];

int main()
{
    int n;
    while(~scanf("%d",&n)&&n)
    {
        //memset(vis,0,sizeof(vis));因为数组太大，所以这里用memset会wa
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&a[i].x,&a[i].y);
            a[i].x+=20000,a[i].y+=20000;//坐标有正有负，而数组下标没有负数 
            vis[a[i].x][a[i].y]=1;
        }
        int ans=0;
        for(int i=0;i<n;i++)
            for(int j=0;j<i;j++)
            {
                if(i==j) continue;
                int x1=a[i].x+a[i].y-a[j].y;
                int y1=a[i].y+a[j].x-a[i].x;
                int x2=a[j].x+a[i].y-a[j].y;
                int y2=a[j].y+a[j].x-a[i].x;
                if(vis[x1][y1]&&vis[x2][y2]) ans++;
                x1=a[i].x+a[j].y-a[i].y;
                y1=a[i].y+a[i].x-a[j].x;
                x2=a[j].x+a[j].y-a[i].y;
                y2=a[j].y+a[i].x-a[j].x;
                if(vis[x1][y1]&&vis[x2][y2]) ans++;
            }
        printf("%d\n",ans/4);
        for(int i=0;i<n;i++) vis[a[i].x][a[i].y]=0;
    }
    return 0;
}