POJ - 1971 E - Parallelogram Counting (数学)

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Parallelogram Counting Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 6969   Accepted: 2446 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 Source Tehran Sharif 2004 Preliminary

题意：n个点，求出这n个点能够组成平行四边形的个数。

分析：

1.  如果有两条不同线段的中点相交，就是一个平行四边形。
2. 求出所有线段中点的集合，求出同个中点的出现的个数k。对于每一个k ，利用组合公式C(k,2)的答案就是平行四边行的个数

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
                   
using namespace std;
			
#define ll long long
 
struct Node{
   int x,y;
}node[1010],mid[2010010];
 
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 f(int k){
    if (k==1) return 0;
    else 
	{
        return k*(k-1)/2;
    }
}
 
int main(){
	int T,n;
    scanf("%d",&T);
    int cas=0;
    while (T--) {
    		
        scanf("%d",&n);
        for (int i=1; i<=n; i++) {
              scanf("%d%d",&(node[i].x),&(node[i].y));
          }
	
        int num=0;
        for (int i=1; i<=n; i++) {
            for (int j=i+1; j<=n; j++) {
                  
                  mid[num].x=node[i].x+node[j].x;
                  mid[num].y=node[i].y+node[j].y;
                  num++;
              }
        }
        sort(mid,mid+num,cmp);
        int k=1;
        ll ans=0;
        for (int i=0; i<num; i++) {
            if (mid[i].x==mid[i+1].x && mid[i].y==mid[i+1].y) 
            	k++;
            else{
                    ans+=f(k);
                    k=1;
                }
        }
       printf("%lld\n",ans); 
       // printf("Case %d: %lld\n",++cas,ans);
 
    }
    return 0;
}