POJ 1972 Parallelogram Counting

Parallelogram Counting

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 5799		Accepted: 1955

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

Source

Tehran Sharif 2004 Preliminary

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

struct node{
	int x;
	int y;
}a[1100];

struct node1//定义中点hash表。//平行四边形的性质，对角线交点为两对角线中点。所以两条线段中点重合就能构成平行四边形。
{
	int x;
	int y;
}zd[1100000];//事实证明用qsort 对int类型排序时间 优于 double类型排序；

int cmp(const void *a,const void *b)
{
     struct node *aa=(node *)a;
     struct node *bb=(node *)b;
	 if((aa->x)!=(bb->x))
		 return((aa->x)-(bb->x));
	 else
		 return((aa->y)-(bb->y));
}

int main()
{
	int t, n;
	int i, j, k;
	int m, sum;

	scanf("%d", &t);
	while(t--)
	{
		scanf("%d", &n);
		for(i=0; i<n; i++)
			scanf("%d%d", &a[i].x, &a[i].y);

		k = 0;
		for(i=0; i<n; i++)
		{
			for(j=i+1; j<n; j++)
			{
				zd[k].x = a[i].x + a[j].x;
				zd[k].y = a[i].y + a[j].y;//之所以不直接存中点坐标是因为排序时间问题。
				k++;
			}
		}
		qsort(zd, k, sizeof(zd[0]), cmp);

		m = 1;
		sum = 0;
		for(i=1; i<k; i++)
		{
			if((zd[i].x==zd[i-1].x) && (zd[i].y==zd[i-1].y))//统计同一中点的线段共有m条，所以能构成m*(m-1)/2个平行四边形。
				m++;
			else
			{
				m = m*(m-1)/2;
				sum += m;
				m = 1;
			}
		}
		printf("%d\n", sum);
	}

	return 0;
}