[概率期望][DP]luogu P3830 随机树

https://www.luogu.org/problem/P3830

 

分析

第一问比较简单，从平均的概念出发，每次扩展我们增加$Average_{i-1}+2$的总深度

所以有$Average_i =\frac{Average_{i-1}*(i-1)+Average_{i-1}+2}{i}$

等于$Average_i =Average_{i-1}+\frac{2}{i}$

第二问作为LaTeX菜菜不想打那么多公示了 挂个网址⑧

https://www.luogu.org/blog/contributation/solution-p3830

 

#include <iostream>
#include <cstdio>
using namespace std;
const int N=110;
int q,n;
double f[N][N],ans;

int main() {
    scanf("%d%d",&q,&n);
    if (q==1) {
        for (int i=2;i<=n;i++) f[i][0]=f[i-1][0]+2.0/i;
        printf("%.6lf",f[n][0]);
    }
    else {
        for (int i=1;i<=n;i++) f[i][0]=1;
        for (int i=2;i<=n;i++)
            for (int j=1;j<i;j++)
                for (int k=1;k<i;k++) f[i][j]+=1.0*(f[k][j-1]+f[i-k][j-1]-f[k][j-1]*f[i-k][j-1])/(i-1);
        for (int i=1;i<=n;i++) ans+=f[n][i];
        printf("%.6lf",ans);
    }
}