CodeForces-28C-Bath Queue-概率DP[ ICPC大连热身D]

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
//前i个澡堂，一共j个人，最长队伍为k的概率
//dp[i][j][k]
double dp[55][55][100];
int a[55];
double C[55][55];
void init() {
    C[0][0] = 1;
    for ( int i = 1; i <= 50; i++ )
    {
        C[i][0] = 1;
        for (int j = 1; j <= i; j++)
            C[i][j] = C[i-1][j-1] + C[i-1][j];
    }
}
int main()
{
    int m,n;
    scanf("%d%d",&n,&m);
    init();//求组合数
    memset(dp,0,sizeof(dp));
    for ( int i = 1; i <= m ; i++ )  scanf("%d",&a[i]);
    dp[0][0][0] = 1; // dp初始化
    for (int i = 1 ; i <= m; i++)
    {
       for (int j = 0; j <= n ; j++)
       {
            for ( int k = 0; k <= n; k++)
            {
                //枚举[i][j][k]
                //计算dp[i][j][k]
                int top = min(k*a[i],j);
                //只有当前澡堂最长队伍小于等于k时。才能转移
                for (int c = 0 ; c <= top ; c++) //i澡堂中人数
                {
                    if (k*a[i]-a[i]+1 > c) // 如果当前 澡堂最大值不是k 只能从dp[i-1][j-c][k]转移
                        dp[i][j][k] += dp[i-1][j-c][k] * pow(i-1,j-c)  / pow(i,j) * C[j][c];
                    else // 如果当前为k，可以从 dp[i-1][j-c][<=K]转移
                         for (int tt = 0; tt<=k ;tt++)
                            dp[i][j][k] += dp[i-1][j-c][tt] * pow(i-1,j-c)  / pow(i,j)* C[j][c];
                }
            }
       }
    }
    double  ans = 0;
    for (int i = 1 ; i <= n; i++)
    {
       ans += dp[m][n][i] * i;
    }
    printf("%.10f\n",ans);
    return 0;
}