\(f(S)\) in the official editorial, the number of trees whose vertex set is \(S\subseteq V\), can be found without Kirchhoff’s theorem. Specifically, it can be calculated by

\[ f(S) = \begin{cases} 1 & (\text{if }|S| \leq 1) \\ \displaystyle \frac{1}{2(|S|-1)} \sum_{T \subset S} f(T) f(S \setminus T) \mathcal{E}(T, S \setminus T) & (\text{otherwise}). \end{cases} \]

Here, for two disjoint sets \(U\) and \(W\) of vertices, \(\mathcal{E}(U, W)\) denotes the number of edges connecting a vertex of \(U\) and a vertex of \(W\), which can be calculated by

\[ \mathcal{E}(U, W) = e(U \cup W) - e(U) - e(W) \]

where \(e(U)\) denotes the number of edges connecting two of the vertex set \(U\).

By precomputing \(e(U)\) for every vertex set \(U\) in a total of \(O(N^2 2^N)\) time, \(f(S)\) can be obtained for every vertex set \(S\) in a total of \(O(3^N)\) time.