poj 1971 Parallelogram Counting

Parallelogram Counting

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 5130		Accepted: 1699

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

#include<iostream>
#include<algorithm>
using namespace std;

struct node
{
    int x;
    int y;
};
node plane[1001];
node middle[500000];

int comp(node a,node b)
{
    if(a.x==b.x)
        return a.y<b.y;
    return a.x<b.x;
}
int main()
{
    int cases;
    int n;
    int i,j,k;
    int sum;
    int total;
    int flag;
//    freopen("E:\\c++\\oj\\t.txt","rt",stdin);
    cin>>cases;
    while(cases--)
    {
        cin>>n;
        for(i=0;i<n;i++)
        {
            cin>>plane[i].x>>plane[i].y;
        }
        k=0;
        for(i=0;i<n-1;i++)
        {
            for(j=i+1;j<n;j++)
            {
                middle[k].x=plane[i].x+plane[j].x;
                middle[k].y=plane[i].y+plane[j].y;
                k++;
            }
        }
        sort(middle,middle+k,comp);
        flag=0;
        total=0;
        sum=1;
        for(i=1;i<k;i++)
        {
            if(middle[i].x==middle[flag].x && middle[i].y==middle[flag].y)
            {
                sum++;
            }
            else
            {
                flag=i;
                total+=sum*(sum-1)/2;
                sum=1;
            }
        }
        cout<<total<<endl;
    }
    return 0;
}