Parallelogram Counting（平行四边形个数，思维转化）

1058 - Parallelogram Counting

Time Limit: 2 second(s)

Memory Limit: 32 MB

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.

Sample Input

Output for Sample Input

0 0

2 0

4 0

1 1

3 1

5 1

-2 -1

8 9

5 7

1 1

4 8

2 0

9 8

Case 1: 5

Case 2: 6

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
",++kase,ans);
    }
    return 0;
}