31 October

https://www.cnblogs.com/RabbitHu/p/51nod1353.html

树形 DP 求所有联通块 (ge K) 的方案数.

切断：(forall iinleft[1, s_u ight]), (displaystyle f(u, i) =sum_{j=1}^{s_v} f(v, j) =f(v, 0)).

不切断：(forall iinleft[1, s_u ight]), (displaystyle sum_{j=1}^{s_v}f(u, i+j) = f(u, i) f(v, j)).

void dfs(int u, int pre) {
	dp[u][1]=sz[u]=1;
	for (int k=head[u], v; k; k=nex[k]) if (v=to[k]!=pre) {
		dfs(v, u);
		for (int i=sz[u]; i; --i) {
			for (int j=1; j<=sz[v]; ++j) dp[u][i+j] = (dp[u][i+j] + dp[u][i]*dp[v][j])%mod;
			dp[u][i]=dp[u][i]*dp[v][0]%mod;
		}
		sz[u]+=sz[v];
	}
	for (int i=K; i<=sz[u]; ++i) dp[u][0]=(dp[u][0]+dp[u][i])%mod;
}

dfs(1, 0);
for (int i=K; i<=sz[i]; ++i) ans=(ans+dp[1][i])%mod;