1058 - Parallelogram Counting

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 next n lines, contains 2 space-separated integersx and y (the coordinates of a point) with magnitude (absolute value) of no more than 1000000000.

Output

For each case, print the case number and the number of parallelograms that can be formed.

题解：

给一系列点，让找到可以组成的平行四边形的个数；

思路：由于平行四边形的交点是唯一的，那么我们只要记录每两个直线的交点，判断交点重复的个数可以求出来了；假设有这个交点有三个重复的，则ans+=3*2/2;

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int INF=0x3f3f3f3f;
typedef long long LL;
const int MAXN=1010;
struct Dot{
    double x,y;
    bool operator < (const Dot &b) const{
        if(x!=b.x)return x<b.x;
        else return y<b.y;
    }
};
Dot dt[MAXN];
Dot num[MAXN*MAXN];
int main(){
    int T,N,kase=0;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&N);
        for(int i=0;i<N;i++)scanf("%lf%lf",&dt[i].x,&dt[i].y);
        int tp=0;
        for(int i=0;i<N;i++)
            for(int j=i+1;j<N;j++){
                if(dt[i].x==dt[j].x&&dt[i].y==dt[j].y)continue;
                num[tp].y=(dt[i].y+dt[j].y)/2;
                num[tp].x=(dt[i].x+dt[j].x)/2;
                tp++;
            }
            sort(num,num+tp);
            LL ans=0,cur=1;
        for(int i=1;i<tp;i++){
            if(num[i].x==num[i-1].x&&num[i].y==num[i-1].y)cur++;
            else{
                ans+=cur*(cur-1)/2;cur=1;
            }
        }
        printf("Case %d: %lld\n",++kase,ans);
    }
    return 0;
}