poj1971:Parallelogram Counting

Description

There are n distinct points in the plane, given by their integer coordinates. Find the number of parallelograms whose vertices lie on these points. In other words, find the number of 4-element subsets of these points that can be written as {A, B, C, D} such that AB || CD, and BC || AD. No four points are in a straight line.

Input

The first line of the input contains a single integer t (1 <= t <= 10), the number of test cases. It is followed by the input data for each test case. 
The first line of each test case contains an integer n (1 <= n <= 1000). Each of the next n lines, contains 2 space-separated integers x and y (the coordinates of a point) with magnitude (absolute value) of no more than 1000000000. 

Output

Output should contain t lines. 
Line i contains an integer showing the number of the parallelograms as described above for test case i. 

Sample Input

2
6
0 0
2 0
4 0
1 1
3 1
5 1
7
-2 -1
8 9
5 7
1 1
4 8
2 0
9 8

Sample Output

5

6

在二维坐标上给出n个点 找出四个点组成一个平行四边形 问你能够组成几个平行四边形

我们知道四边形对角线相较于中点 而中点的横纵坐标就等于对角线两点横纵坐标和的一半 这样我们只要找出中点相同的边的数量

然后两两组合即可

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct node
{
	int x,y;
}dian[10001];
node mid[500001];
bool cmp(node a,node b)
{
	if(a.x !=b.x )
	return a.x <b.x ;
	else
	return a.y <b.y ;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n;
		scanf("%d",&n);
		for(int i=0;i<n;i++)
		{
		 scanf("%d%d",&dian[i].x ,&dian[i].y );
		}
		int cut=0;
      for(int i=0;i<n;i++)
      {
      	for(int j=i+1;j<n;j++)//把给出的点两两组合连成边 
      	{
      		 mid[cut].x =dian[i].x +dian[j].x ;//这里没有除以2是为了避免出现小数 结果是不变的 
      		 mid[cut].y =dian[i].y+dian[j].y ;
      		 cut++;
		  }
	  }
	  sort(mid,mid+cut,cmp); //排序 让中点坐标相同的排列在一起 
	  int ans=0;
	  int flog=0;//起点边的坐标 
	  int sum=1;//每次在新区间开始的时候中点数为1(起点边的中点)   s 
	  for(int i=1;i<cut;i++)
	  {
	     if(mid[flog].x ==mid[i].x&&mid[flog].y==mid[i].y )
		 {
		 	sum++;
			 }	
			 else
			 {
			 	ans=ans+sum*(sum-1)/2;//找出的边两两组合 
			 	sum=1;//初始化
			 	flog=i;//更新新的起点 
			 }
	  }
	  printf("%d\n",ans);
	}
	return 0;
 }