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
题意： 计算平行四边形的个数，，，，，
思路：用结构体记录任意两个点之间的x坐标和与y坐标和，(相当于这两个点作为一条对角线的情况），然后找出对角线相交于同一点的个数k（结构体中的x和y相等），然后C(k,2)就OK,因为若多条对角线相交于同一个中点，则任意选择两条就可以构成一个平行四边形；
代码：
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int x[1001]; int y[1001]; struct node { int x; int y; }a[1000*1000/2]; int cmp(node a,node b) { if(a.x!=b.x) { return a.x<b.x; } else return a.y<b.y; } int main() { int t,n,mm=1; scanf("%d",&t); while(t--) { scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%d%d",&x[i],&y[i]); } int t=0; for(int i=0;i<n;i++) { for(int j=i+1;j<n;j++) { a[t].x=x[i]+x[j]; a[t].y=y[i]+y[j]; t++; } } int num=0,ans=1,cnt=0; sort(a,a+t,cmp); for(int i=1;i<t;i++) { if(a[i].x==a[num].x&&a[i].y==a[num].y) { ans++;//记录经过一个中点的对角线的个数； } else { cnt+=ans*(ans-1)/2;//计算经过这个中点的平行四边形的个数； num=i;//记录经过下一个中点的一条对角线，作为后面比较的对象； ans=1;//每次计算经过不同中点的对角线个数时，都要初始化为1； } } printf("Case %d: %d\n",mm++,cnt); } return 0; }
