題面

luogu

題解

第一問:

設\(f[i]\)表示\(i\)步操作後,平均深度期望

\(f[i] = \frac {f[i - 1] * (i - 1)+f[i-1]+2}{i}=f[i-1]+\frac{2}{i}\)

第二問就比較難受了:

\(E(x)=∑_{i=1}^{x}P\)

隨機變量\(x\)的期望為對於所有\(i\)，\(i≤x\)的概率之和

我們設\(f[i][j]\)表示\(i\)步後,樹的深度\(>=j\)的概率

我們每次新建一個根,然後枚舉左右子樹分配節點情況

\(f[i][j] = \frac{1}{i-1}\sum_{k=1}^{i-1}(f[k][j-1]*1+f[i-k][j-1]*1-f[k][j-1]*f[i-k][j-1])\)

然後

\(Ans = \sum_{i=1}^{n-1}f[n][i]\)

Code

#include<bits/stdc++.h>

#define LL long long
#define RG register

using namespace std;
template<class T> inline void read(T &x) {
    x = 0; RG char c = getchar(); bool f = 0;
    while (c != '-' && (c < '0' || c > '9')) c = getchar(); if (c == '-') c = getchar(), f = 1;
    while (c >= '0' && c <= '9') x = x*10+c-48, c = getchar();
    x = f ? -x : x;
    return ;
}
template<class T> inline void write(T x) {
    if (!x) {putchar(48);return ;}
    if (x < 0) x = -x, putchar('-');
    int len = -1, z[20]; while (x > 0) z[++len] = x%10, x /= 10;
    for (RG int i = len; i >= 0; i--) putchar(z[i]+48);return ;
}
const int N = 110;
int q, n;
namespace cpp1 {
    double f[N];
    void main() {
        for (int i = 2; i <= n; i++) f[i] = f[i - 1] + 2.0 / i;
        printf("%lf\n", f[n]);
        return ;
    }
}
namespace cpp2 {
    double f[N][N];
    void main() {
        for (int i = 1; i <= n; i++) f[i][0] = 1;
        for (int i = 2; i <= n; i++)
            for (int j = 1; j < n; j++) {
                for (int k = 1; k < i; k++)
                    f[i][j] += f[k][j - 1] + f[i - k][j - 1] - f[k][j - 1] * f[i - k][j - 1];
                f[i][j] /= (i - 1);
            }
        double ans = 0;
        for (int i = 1; i < n; i++) ans += f[n][i];
        printf("%lf\n", ans);
    }   
}

int main() {
    read(q), read(n);
    if (q == 1) cpp1 :: main();
    else cpp2 :: main();
    return 0;
}
 

洛谷P3830 [SHOI2012]隨機樹(期望dp)