poj Squares 2002 （STL&&枚举+转换） 统计正方形的个数

Squares

Time Limit: 3500MS Memory Limit: 65536KTotal Submissions: 17886 Accepted: 6845

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

//题意：

给你n个点，让你求出这n个点能组成几个正方形。

参考博客：http://m.blog.csdn.net/blog/u013480600/39500427

#include<stdio.h>#include<string.h>#include<algorithm>#define N 1010using namespace std;struct zz{int x;int y;friend bool operator<(zz a,zz b){return a.x<b.x||(a.x==b.x&&a.y<b.y);}}q[N<<1],tmp;int main(){int n,i,j,k;while(scanf("%d",&n),n){for(i=0;i<n;i++)scanf("%d%d",&q[i].x,&q[i].y);sort(q,q+n);int cnt=0;for(i=0;i<n;i++){for(j=i+1;j<n;j++){tmp.x=q[i].x+q[i].y-q[j].y;tmp.y=q[i].y+q[j].x-q[i].x;if(!binary_search(q,q+n,tmp))continue;tmp.x=q[j].x+q[i].y-q[j].y;tmp.y=q[j].y+q[j].x-q[i].x;if(!binary_search(q,q+n,tmp))continue;cnt++;}}printf("%d\n",cnt/2);}return 0;}