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;
}
转载于:https://www.cnblogs.com/sssy/p/9520637.html
