【LightOJ - 1058】Parallelogram Counting

水深无声 2022-09-25 05:22 154阅读 0赞

[Parallelogram Counting][]

**Time Limit:**2000MS **Memory Limit:**32768KB **64bit IO Format:**%lld & %llu

Submit [Status][] [Practice][] [LightOJ 1058][] [![12x12_uDebug.png][]uDebug][12x12_uDebug.png_uDebug]

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

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 integers **x** 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

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

Sample Output

Case 1: 5

Case 2: 6

根据平行四边形性质，两对角线中点一定重合，所以先取任意两点间的中点，然后在对这些中点排序sort一下，然后用组合数公式C（n，2）求出任意相同的点中任意两点组合的个数求和即可。

#include<cstdio>
    #include<iostream>
    #include<algorithm>
    #include<cstring>
    using namespace std;
    const int N = 1005;
    struct node {
    	int x;
    	int y;
    } s[N],mid[N*N];
    bool cmp(node A,node B) {
    	if(A.x==B.x)
    		return A.y<B.y;
    	return A.x<B.x;
    }
    int main() {
    	int T,p=0;
    	scanf("%d",&T);
    	while(T--) {
    		int n;
    		scanf("%d",&n);
    		for(int i=0; i<n; i++) {
    			scanf("%d%d",&s[i].x,&s[i].y);
    		}
    		int cnt=0;
    		for(int i=0; i<n; i++) {
    			for(int j=i+1; j<n; j++) {
    				mid[cnt].x=s[i].x+s[j].x;//不能除二，可能为奇数导致误差 
    				mid[cnt].y=s[i].y+s[j].y;
    				cnt++;
    			}
    		}
    		sort(mid,mid+cnt,cmp);
    		int count=1,sum=0;
    		for(int i=0; i<cnt;i++) {
    			if(mid[i].x==mid[i+1].x&&mid[i].y==mid[i+1].y) {
    				count++;
    			} else {
    				sum=sum+count*(count-1)/2;
    				count=1;
    			}
    		}
    		printf("Case %d: %d\n",++p,sum);
    	}
    	return 0;
    }

[Parallelogram Counting]: http://acm.hust.edu.cn/vjudge/problem/visitOriginUrl.action?id=26775
[Status]: http://acm.hust.edu.cn/vjudge/contest/127946#status//H/0
[Practice]: http://acm.hust.edu.cn/vjudge/problem/26775
[LightOJ 1058]: http://acm.hust.edu.cn/vjudge/problem/26775/origin
[12x12_uDebug.png]: http://acm.hust.edu.cn/vjudge/images/12x12_uDebug.png
[12x12_uDebug.png_uDebug]: https://www.udebug.com/LOJ/1058