POJ-2002 Squares解题报告

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：就是给你N个不重复的点，找出所有的不同的正方形。

题目链接：http://poj.org/problem?id=2002

方法：排序及二分查找

思路：对于点，我们可以定义一个结构体，来联系点的坐标，然后进行点升序排列，方便我们后面二分查找，我们对于每两个点来构造一个正方形的边，我们可以想象一下，在平面坐标系里面，一个正方形始终有一个边垂直X轴或者与x轴正方向成锐角，我们就把构造的边当做正方形的这种边来计算正方形的其余两点，这样就不会有重复的结果，而且不需要分情况，求其余两点坐标很好求，画图用几何算出来，求出来后进行二分查找，判断是否存在，不二分查找容易超时。

算法实现：

 
#include<cstdio>#include<iostream>#include<algorithm>using namespace std;struct point{int x;int y;};point a[1003];int cmp(const void *a,const void *b){if( ((point*)a)->x!=((point*)b)->x )return ((point*)a)->x-((point*)b)->x;elsereturn ((point*)a)->y-((point*)b)->y;}bool search(point p,int n){int low,high,mid;low=0;high=n-1;while(high-low>=0){mid=(low+high)/2;if(cmp(&a[mid],&p)>0)high=mid-1;else if(cmp(&a[mid],&p)==0){return true;}elselow=mid+1;}return false;}int main(){int n,i,j,sum;point b,c;while(scanf("%d",&n)!=EOF){if(n==0)break;for(i=0;i<n;i++){scanf("%d%d",&a[i].x,&a[i].y);}sum=0;qsort(a,n,sizeof(a[0]),cmp);for(i=0;i<n;i++)for(j=i+1;j<n;j++){if(a[j].y>a[i].y){int dx=a[j].x-a[i].x;int dy=a[j].y-a[i].y;b.x=a[i].x-dy;b.y=a[i].y+dx;c.x=a[j].x-dy;c.y=a[j].y+dx;if( search(b,n) && search(c,n)){sum++;}}}printf("%d\n",sum);}return 0;    }