洛谷P3830 [SHOI2012]随机树(期望dp)

题面

luogu

题解

第一问:

设(f[i])表示(i)步操作后,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})

第二问就比较难受了:

(E(x)=∑_{i=1}^{x}P)

随机变量(x)的期望为对于所有(i)，(i≤x)的概率之和

我们设(f[i][j])表示(i)步后,树的深度(>=j)的概率

我们每次新建一个根,然后枚举左右子树分配节点情况

(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]))

然后

(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
", 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
", ans);
	}	
}

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