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
题意:在二维平面上,给你n个点的坐标,求任取四点构成正方形的个数
题解:好吧,暴力O(n^4)又t了……
那么折半枚举吧
枚举两个点连成一条线,然后根据这条线可以得到正方形的另外两个点,根据hash可以快速确定点是否在n个点中,即可以计算出正方形的个数
代码如下:
#include<map>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x3f3f3f3f
using namespace std; struct node
{
int x,y;
} a[]; vector<long long> g[];
int n; int cmp(node x,node y)
{
if(x.x==y.x)
{
return x.y<y.y;
}
return x.x<y.x;
} int main()
{ while(scanf("%d",&n)==&&n)
{
int ans=;
vector<long long> g[];
for(int i=; i<=n; i++)
{
scanf("%d%d",&a[i].x,&a[i].y);
}
sort(a+,a+n+,cmp);
for(int i=; i<=n; i++)
{
long long key=(a[i].x*a[i].x+a[i].y*a[i].y)%;
g[key].push_back(i);
}
for(int i=; i<=n; i++)
{
for(int j=i+; j<=n; j++)
{
int x1,y1,x2,y2;
int dx=(a[j].x-a[i].x),dy=(a[j].y-a[i].y);
// if(dx<0||dy<0)
// {
// continue;
// }
x1=a[i].x-dy;
y1=a[i].y+dx;
x2=a[j].x-dy;
y2=a[j].y+dx;
int flag=;
long long key=(x1*x1+y1*y1)%;
for(int k=; k<g[key].size(); k++)
{
if(a[g[key][k]].x==x1&&a[g[key][k]].y==y1)
{
flag+=;
}
}
key=(x2*x2+y2*y2)%;
for(int k=; k<g[key].size(); k++)
{
if(a[g[key][k]].x==x2&&a[g[key][k]].y==y2)
{
flag+=;
}
}
if(flag==)
{
// if(a[i].x==x1&&a[i].y==y1||a[j].x==x2&&a[j].y==y2||a[i].x==x2&&a[i].y==y2||a[j].x==x1&&a[j].y==y1)
// {
// continue;
// }
// printf("%d %d\n",dx,dy);
// printf("%d %d %d %d %d %d %d %d\n",a[i].x,a[i].y,a[j].x,a[j].y,x1,y1,x2,y2);
ans++;
}
}
}
printf("%d\n",ans/);
} }


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