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 nextn lines, contains2 space-separated integers x and y (the coordinates of a point) with magnitude (absolute value) of no more than1000000000.
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
利用平行四边形的性质，两对角线的中点相等来做，可以把每两个点的中点都求出来，然后排序找相等的点，相加即可；
#include <iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<algorithm> #define INF 0xfffffff using namespace std; struct point { int x,y; }; point p[1100],a[500100]; int cmp(point a,point b) { if(a.x!=b.x) { return a.x<b.x; } else { return a.y<b.y; } } int main() { int t,n,i,j,ans,count; int T=0; scanf("%d",&t); while(t--) { memset(p,0,sizeof(p)); memset(a,0,sizeof(a)); scanf("%d",&n); for(i=0; i<n; i++) scanf("%d%d",&p[i].x,&p[i].y); int k=0; for(i=0; i<n; i++) for(j=i+1; j<n; j++) { a[k].x=1.0*(p[i].x+p[j].x); a[k].y=1.0*(p[i].y+p[j].y); k++; } sort(a,a+k,cmp);//这里注意排序要写快排，用冒泡会超时，快排应该用sort，用c语言的qsort无法进行正确的排序 ans=1; count=0; for(i=0; i<k-1; i++) { if(a[i].x==a[i+1].x&&a[i].y==a[i+1].y) ans++; else { count+=(ans-1)*ans/2; ans=1; } } count+=(ans-1)*ans/2; printf("Case %d: %d\n",++T,count); } return 0; }