lightoj1058 && poj1791 求平行四边形个数

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最新推荐文章于 2022-01-06 20:20:41 发布

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分类专栏：基础题 Lightoj poj

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该博客介绍了如何解决一道编程题，题目要求找出一组给定坐标点能构成的平行四边形个数。通过利用平行四边形对角线中点相等的性质，计算每对点的中点并排序，查找重复的中点来确定平行四边形的数量。给出的样例输入和输出展示了测试用例的处理方式。

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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 nextn lines, contains2 space-separated integers x and y (the coordinates of a point) with magnitude (absolute value) of no more than1000000000.

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

利用平行四边形的性质，两对角线的中点相等来做，可以把每两个点的中点都求出来，然后排序找相等的点，相加即可；

#include <iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#define INF 0xfffffff
using namespace std;
struct point
{
    int x,y;
};
point p[1100],a[500100];
int cmp(point a,point b)
{
    if(a.x!=b.x)
    {
        return a.x<b.x;
    }
    else
    {
        return a.y<b.y;
    }
}

int main()
{
    int t,n,i,j,ans,count;
    int T=0;
    scanf("%d",&t);
    while(t--)
    {
        memset(p,0,sizeof(p));
        memset(a,0,sizeof(a));
        scanf("%d",&n);
        for(i=0; i<n; i++)
            scanf("%d%d",&p[i].x,&p[i].y);
        int k=0;
        for(i=0; i<n; i++)
            for(j=i+1; j<n; j++)
            {
                a[k].x=1.0*(p[i].x+p[j].x);
                a[k].y=1.0*(p[i].y+p[j].y);
                k++;
            }
        sort(a,a+k,cmp);//这里注意排序要写快排，用冒泡会超时，快排应该用sort，用c语言的qsort无法进行正确的排序
        ans=1;
        count=0;
        for(i=0; i<k-1; i++)
        {
            if(a[i].x==a[i+1].x&&a[i].y==a[i+1].y)
                ans++;
            else
            {
                count+=(ans-1)*ans/2;
                ans=1;
            }
        }
            count+=(ans-1)*ans/2;
        printf("Case %d: %d\n",++T,count);
    }
    return 0;
}