杂题
题意
\(n\)个点的无向图,贡献为连通块个数的\(m\)次方,多次查询\(n\le 3\times 10^4,m\le 15\)
做法一
令\(\hat {F(x)}\)为无向连通图个数\(EGF\)
令\(\hat {G_m(x)}\)为答案的\(EGF\)
有:\(\hat{G_m(x)}=\sum \frac{\hat{F(x)}^i\times i^m}{i!}\)
显然:\(\hat{G_m(x)}=\hat{F(x)}\frac{d\hat{G_{m-1}(x)}}{d\hat{F(x)}}\),边界\(\hat{G_0(x)}=e^{\hat{F(x)}}\)
\(\frac{d\hat{G_{m-1}(x)}}{d\hat{F(x)}}=\frac{d\hat{G_{m-1}(x)}}{dx}\cdot \frac{dx}{d\hat{F(x)}}\)
做法二
将\(x^m\)转成下降幂:\(x^m=\sum S_{m,i}x^{\underline i}\)
考虑组合意义,\(x^{\underline i}\)为有序选择\(i\)个位置
令\(f_{i,n}\)为\(n\)个点的有序选择\(i\)个位置的方案数,令\(g_i\)为\(i\)个点的无向联通图个数
\(f_{i,n}=\sum\limits_{k=1}^n {n\choose k}g_kf_{i-1,n-k}\)
\(g_i\)可以用\(ln\)求