POJ2002 Squares

原题链接：http://poj.org/problem?id=2002
博主的中文题面：
https://www.luogu.org/problemnew/show/T23926

Squares

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

给一堆点，求能构成多少正方形

题解

n个点可以构成n^2/2条边，对于每一条边，都可以查找对应的可以构成正方形边，这个过程可以二分查找，也可以用哈希，因为每个正方形会被两组对边统计两次，所以需要除以二。

代码

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int M=1005;
struct pt{int x,y;};
bool operator == (const pt &a,const pt &b){return a.x==b.x&&a.y==b.y;}
bool operator < (const pt &a,const pt &b){return a.x!=b.x?a.x<b.x:a.y<b.y;}
pt p[M];
int n;
void in()
{for(int i=1;i<=n;++i)scanf("%d%d",&p[i].x,&p[i].y);
}
int find(int le,int ri,pt x)
{int mid=(le+ri)>>1;if(p[mid]==x) return mid;if(le==ri) return -1;if(p[mid]<x) return find(mid+1,ri,x);return find(le,mid,x);
}
void ac()
{long long ans=0;sort(p+1,p+1+n);pt f;for(int i=1;i<=n;++i)for(int j=i+1;j<=n;++j){f.x=p[i].x-p[j].y+p[i].y;f.y=p[i].y+p[j].x-p[i].x;if(find(1,n,f)<0) continue;f.x=p[j].x-p[j].y+p[i].y;f.y=p[j].y+p[j].x-p[i].x;if(find(1,n,f)<0) continue;ans++;}printf("%lld\n",ans/2);
}
void reset()
{memset(p,0,sizeof(p));
}
int main()
{while(scanf("%d",&n)&&n){reset();in();ac();}return 0;
}

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