POJ 2002 Squares

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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

Source

Rocky Mountain 2004

/*
题目大意是在二维平面坐标中给定N个点，然后通过查找的方式找出其中有多少个正方形。

解决方法用的是STL中的set函数，用pair将坐标作为一个数据类型存放在set中，然后每次枚举两个点，判断正方形的其他两个点是否存在。
*/

#include <iostream>#include <algorithm>#include <cstring>#include <string>#include <stack>#include <cmath>#include <cstdio>#include <vector>#include <queue>#include <set>using namespace std;struct node{    int x,y;}a[1010];int f(int x1,int y1){    return (x1+y1)/2;}int main(){    int n;    while(~scanf("%d",&n) && n)    {        set<pair<int,int> >s;        for(int i=0;i<n;i++)        {            scanf("%d %d",&a[i].x,&a[i].y);            a[i].x*=2;a[i].y*=2; //将每个点*2处理，防止之后的浮点数            s.insert(pair<int,int>(a[i].x,a[i].y)); //存入set        }        int sum = 0;        for(int i=0;i<n;i++)        {            for(int j=i+1;j<n;j++)            {                int x = f(a[i].x,a[j].x); //求中点横坐标                int y = f(a[i].y,a[j].y); //求中点竖坐标                if(s.find(pair<int,int>(x+(-1*(a[i].y-y)),y+(a[i].x-x)))!=s.end() && s.find(pair<int,int>(x-(-1*(a[i].y-y)),y-(a[i].x-x)))!=s.end()) //判断两个点是否存在                {                    sum++;                }            }        }        printf("%d\n",sum/2);//因为枚举两个点的时候，正方形有两个对角线，每个正方形会遇到两次，所以除以2    }    return 0;}

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