1971 Parallelogram Counting 平面点集能组成多少个平行四边形 中点排序

Parallelogram Counting

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 3709		Accepted: 1149

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

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
bool equal(double x,double y)
{
    return fabs(x-y)<1e-4;
}
struct _Point
{
    double x,y;
};
_Point p[1100];
_Point a[1100*1100];
bool cmp(_Point h,_Point k)
{
    if(h.x!=k.x) return h.x<k.x;
    return h.y<k.y;
}
int main()
{
    int ci;scanf("%d",&ci);
    while(ci--)
    {
        int n;scanf("%d",&n);
        for(int i=0;i<n;i++) scanf("%lf%lf",&p[i].x,&p[i].y);
        int pl=0;
        for(int i=0;i<n;i++) for(int j=i+1;j<n;j++) a[pl].x=(p[i].x+p[j].x)/2,a[pl++].y=(p[i].y+p[j].y)/2;
        //对于一个平行四边形，其两条对角线的的交点是其中点。因此可以计算每条线段的中点是否相等判断是否相等
        sort(a,a+pl,cmp);//中点排序
        long long cnt=0,t=0;
        for(int i=0;i<pl;i++)
        {
            if(i&&!(equal(a[i].x,a[i-1].x)&&equal(a[i].y,a[i-1].y)))
            {
                cnt+=t*(t-1)/2;
                t=1;
            }
            else t++;
        }
        cout<<cnt<<endl;
    }
    return 0;
}