POJ 1971 Parallelogram Counting

Parallelogram Counting

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 5605		Accepted: 1885

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
题目大意：给定n个点的坐标，问n个点能够组成多少个平行四边形。

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <stdlib.h>
using namespace std;

typedef struct _point
{
    int x;
    int y;
}_Point;

_Point p[1000100];
_Point temp[1010];

bool cmp(const _Point& point1, const _Point& point2)
{
    if (point1.x == point2.x)
    {
        return point1.y > point2.y;
    }
    else
    {
        return point1.x > point2.x;
    }
}

int main()
{
    int N;
    scanf("%d", &N);
    while(N--)
    {
        int n;
        scanf("%d", &n);
        int nCount = 0;
        for (int i = 0; i < n; i++)
        {
            scanf("%d%d", &temp[i].x, &temp[i].y);
        }
        for (int i = 0; i < n; i++)
        {
            for (int j = i + 1; j < n; j++)
            {
                p[nCount].x = temp[i].x + temp[j].x;
                p[nCount++].y = temp[i].y + temp[j].y;
            }
        }
        sort(p, p + nCount, cmp);
        int k = 1;
        int sum = 0;
        for (int i = 0; i < nCount; i++)
        {
            if (p[i].x == p[i + 1].x && p[i].y == p[i + 1].y)
            {
                k++;
            }
            else
            {
                sum += (k - 1) * k / 2;
                k = 1;
            }
        }
        printf("%d
", sum);
    }
    return 0;
}