POJ1791 Parallelogram Counting[数学题平行四边形求个数]

Parallelogram Counting

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 5361		Accepted: 1794

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

 

 

 

 

【题目大意】：给出n个点，求出这n个点能够组成平行四边形的个数。

【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，就是一个平行四边形

2）利用点坐标求出中点的集合，离散化后求出同个中点的出现的个数k。
3）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

 

 

code:

#include<iostream>
#include<algorithm>
using namespace std;

#define MAXN 1010

typedef struct point
{
    int x,y;
}Point;
Point point[MAXN];
Point mid[MAXN*MAXN];

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

int main()
{
    int t;
    int i,j;
    int sum;
    int n;
    int cnt;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        sum=0;
        cnt=0;
        for(i=0;i<n;i++)
            scanf("%d%d",&point[i].x,&point[i].y);
        for(i=0;i<n;i++)
            for(j=i+1;j<n;j++)
            {
                mid[cnt].x=(point[i].x+point[j].x);
                mid[cnt].y=(point[i].y+point[j].y);
                cnt++;
            }
        sort(mid,mid+cnt,cmp);
        int count=1;
        for(i=0;i<cnt;i++)
        {
            if(mid[i].x==mid[i+1].x&&mid[i].y==mid[i+1].y)
                count++;
            else
            {
                sum+=(count-1)*count/2;
                count=1;
            }
        }
        printf("%d\n",sum);
    }
    return 0;
}