POJ

Parallelogram Counting

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 6466 Accepted: 2248

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

260 02 04 01 13 15 17-2 -18 95 71 14 82 09 8

Sample Output

56 当两条线段中点重合时，四个顶点组成一个平行四边形
#include<iostream>#include<algorithm>using namespace std;struct AP{    int x,y;} a[1008];struct AD{    double x,y;} com[510000];bool compare(AD a,AD b);int main(){    int N;    cin>>N;    while(N--)    {        int n,x=0;        cin>>n;        for(int i=0; i<n; i++)            cin>>a[i].x>>a[i].y;        for(int i=0; i<n-1; i++)            for(int j=i+1; j<n; j++)            {                com[x].x=(a[i].x+a[j].x)/2.0;                com[x].y=(a[i].y+a[j].y)/2.0;                x++;            }        sort(com,com+x,compare);        double comx=com[0].x,comy=com[0].y;        int ans=0,add=1;        for(int i=1; i<x; i++)             if(comx==com[i].x&&comy==com[i].y)                add++;            else            {                comx=com[i].x;                comy=com[i].y;                ans+=(add-1)*add/2;                add=1;            }        ans+=(add-1)*add/2;        cout<<ans<<endl;    }    return 0;}bool compare(AD a,AD b){    if(a.x!=b.x)        return a.x<b.x;    else        return a.y<b.y;}