bzoj2830: [Shoi2012]随机树

题目链接

bzoj2830: [Shoi2012]随机树

题解

q1好做

设f[n]为扩展n次后的平均深度

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

化简之后也就是\(f[n] = f[n - 1] + \frac{2}{n}\)

q2也好做

设f[i][j]表示扩展i次，树高为j的概率，对于左右儿子，子问题显然是一样的

枚举左右子树的i j 转移

\(f[i][std::max(l,k) + 1] += f[j][k] * f[i - j][l] / (i - 1)\)

\(n^4\)可过

代码

#include<cstdio>
#include<algorithm> 

inline int read() {
	int x = 0,f = 1;
	char c = getchar();
	while(c < '0' || c > '9')c = getchar();
	while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = getchar();
	return x * f;
} 

const int maxn = 1007;
int type,n;
double f[maxn];
void solve1() {
	f[0] = 0 ;
	for(int i = 2;i <= n;++ i)
		f[i] = f[i - 1] + 2.0 / i;
	printf("%lf\n",f[n]);
}
double F[maxn][maxn];
void solve2() {
	F[1][0] = 1.0;
	for(int i = 2;i <= n;++ i)
		for(int j = 1;j < i;++ j) {
			for(int k = 0;k <= j;++ k)
				for(int l = 0;l <= (i - j);++ l) {
					F[i][std::max(l,k) + 1] += F[j][k] * F[i - j][l] / (i - 1);
				}
		}
	double ans = 0;
	for(int i = 1;i <= n;++ i) ans += i * F[n][i];
	printf("%lf\n",ans);
}
int main() {
	type = read(),n = read();
	type == 1 ?  solve1() : solve2();
	return 0;
}