【Uva11806】Cheerleaders

最新推荐文章于 2019-08-20 15:24:00 发布

清疚

最新推荐文章于 2019-08-20 15:24:00 发布

阅读量422

点赞数

CC 4.0 BY-SA版权

分类专栏：题解数论文章标签： c++ UVA

本文链接：https://blog.youkuaiyun.com/qq_38064038/article/details/78221626

题解同时被 2 个专栏收录

56 篇文章

订阅专栏

数论

13 篇文章

订阅专栏

本文探讨了一个关于体育赛事中啦啦队布局的问题。在一个矩形网格内放置一定数量的啦啦队成员，确保每边至少有一名成员，同时所有成员都被合理安排。采用组合数学的方法结合容斥原理来解决该问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题面

In most professional sporting events, cheerleaders play a major role in entertaining the spectators. Their roles are substantial during breaks and prior to start of play. The world cup soccer is no exception. Usually the cheerleaders form a group and perform at the centre of the field. In addition to this group, some of them are placed outside the side line so they are closer to the spectators. The organizers would like to ensure that at least one cheerleader is located on each of the four sides. For this problem, we will model the playing ground as an M × N rectangular grid. The constraints for placing cheerleaders are described below:
• There should be at least one cheerleader on each of the four sides. Note that, placing a cheerleader on a corner cell would cover two sides simultaneously.
• There can be at most one cheerleader in a cell.
• All the cheerleaders available must be assigned to a cell. That is, none of them can be left out.

The organizers would like to know, how many ways they can place the cheerleaders while maintaining the above constraints. Two placements are different, if there is at least one cell which contains a cheerleader in one of the placement but not in the other.

题意

在一个 $m$ 行 $n$ 列的矩形网格里放 $k$ 个石子，问有多少种放法？每个格子里最多放一个石子，所有石子都要用完，并且第一行，最后一行，第一列，最后一列都得有石子。

解法

组合数+容斥：
首先不考虑约束条件，那么我们很容易得出答案为： $C_{n*m}^k$
直接解决这个问题比较难，所以考虑用总方案-不合法的方案，现在重点就变为了求不合法的方案：
设 $A$ 为第一排不放石子的方案集合， $B$ 为第 $m$ 排不放石子的方案集合， $C$ 为第一列不放石子的方案集合， $D$ 为最后一列不放石子的方案集合，显然， $A∩B∩C∩D≠∅$ ，那么就需要使用容斥
因为有4种初始状态，所以可以得出共有16种情况，考虑使用一个四位二进制数（即0~15）来表示这些情况，所以有：
$ans=C_{n*m}^k-（|A|+|B|+|C|+|D|）+（|A∩B|+|A∩C|+|A∩D|+|B∩C|+|B∩D|+|C∩D|）-（|A∩B∩C|+|A∩B∩D|+|A∩C∩D|+|B∩C∩D|+|A∩B∩C∩D|）$
剩下的就请看代码

复杂度

O（ $1000^2+T*16$ ）

代码

#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#define Lint long long int
using namespace std;
const int M=1000007;
const int MAXN=1010;
int C[MAXN][MAXN];
int T,n,m,k;
int ans;
void Prepare()
{
    C[0][0]=1;
    for(int i=1;i<=400;i++)
    {
        C[i][0]=1;
        for(int j=1;j<=i;j++)   C[i][j]=(C[i-1][j-1]+C[i-1][j])%M;
    }
}
int main()
{
    int Case=0,r,c;
    Prepare();
    scanf("%d",&T);
    while( T-- )
    {
        ans=0;
        scanf("%d%d%d",&n,&m,&k);
        for(int s=0,b;s<=15;s++)
        {
            r=n,c=m,b=0;
            if( s&1 )   r--,b++;
            if( s&2 )   r--,b++;
            if( s&4 )   c--,b++;
            if( s&8 )   c--,b++;
            if( b&1 )   ans=(ans+M-C[r*c][k])%M;
            else   ans=(ans+C[r*c][k])%M;
        }
        printf("Case %d: %d\n",++Case,ans);
    }
    return 0;
}