【解题思路】：

1）平行四边形的对角线的中点一定相交。<=> 如果有两条不同线段的中点相交࿰c;就是一个平行四边形

【代码】：

<code class="language-cpp">#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <cmath>
#include <string>
#include <cctype>
#include <map>
#include <iomanip>
                   
using namespace std;
                   
#define eps 1e-8
#define pi acos(-1.0)
#define inf 1<<30
#define pb push_back
#define lc(x) (x << 1)
#define rc(x) (x << 1 | 1)
#define lowbit(x) (x & (-x))
#define ll long long

struct Node{
   int x,y;
}node[1010],mid[1001000];
int T,n,summ,num;

int cmp(const Node &a,const Node &b){
    if (a.x==b.x) return a.y<b.y; else return a.x<b.x;
}

int count(int k){
    if (k==1) return 0;
    else {
        return k*(k-1)/2;
    }
}

int main(){
    cin >> T;
    while (T--) {
        summ=0;
        scanf("%d",&n);
        for (int i=1; i<=n; i++) {
              scanf("%d%d",&(node[i].x),&(node[i].y));
          }
        num=-1;
        for (int i=1; i<=n; i++) {
            for (int j=i+1; j<=n; j++) {
                  num++;
                  mid[num].x=node[i].x+node[j].x;
                  mid[num].y=node[i].y+node[j].y;
              }
        }
        sort(mid,mid+num+1,cmp);
        Node temp;
        int k;
        k=1;
        for (int i=0; i<=num; i++) {
            if (mid[i].x==mid[i+1].x && mid[i].y==mid[i+1].y) k++;
            else {
                    summ+=count(k);
                    k=1;
                }
        }
        cout << summ << endl;

    }
    return 0;
}

code>