POJ 1791 Parallelogram Counting（求平行四边形数量）

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

Input starts with an integer T (≤ 15), denoting the number of test cases.

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

For each case, print the case number and the number of parallelograms that can be formed.

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

Case 1: 5

Case 2: 6

给出点的坐标求出能连成平行四边形的数量。

思路：

　　用结构体记录每两个点的x坐标和与y坐标和（ 相当于记录这两个点为对角线的情况 ）然后循环判断多少个对角线的被交点平分（结构体中x，y的值相等）求出对角线的数量，然后C（sum，2）。

 #include<cstdio>
 #include<algorithm>
 using namespace std;
 int x[],y[];
 struct stu
 {
     int x,y;
 }st[*/];
 bool cmp(stu a,stu b)
 {
     if(a.x != b.x)
         return a.x < b.x;
     else
         return a.y < b.y;
 }
 int main()
 {
     int t,ss=;
     scanf("%d",&t);
     while(t--)
     {
         int n,i,j;
         scanf("%d",&n);
         for(i =  ; i < n ; i++)
         {
             scanf("%d %d",&x[i],&y[i]);
         }
         int k=;
         for(i =  ; i < n ; i++)
         {
             for(j = i+ ; j < n ; j++)
             {
                 st[k].x=x[i]+x[j];
                 st[k++].y=y[i]+y[j];
             }
         }
         sort(st,st+k,cmp);
         int num=,sum=,ans=;
         for(i =  ; i < k ; i++)
         {
             if(st[num].x == st[i].x && st[num].y == st[i].y)
                 sum++;                            //sum代表线的数量
             else
             {
                 ans+=(sum*(sum-)/);            //这里是C(sum,2)
                 num=i;
                 sum=;

             }
         }
         printf("Case %d: %d\n",++ss,ans);
     }
 }