loj2542 「PKUWC2018」随机游走 【树形dp + 状压dp + 数学】

题目链接

loj2542

题解

设\(f[i][S]\)表示从\(i\)节点出发，走完\(S\)集合中的点的期望步数

记\(de[i]\)为\(i\)的度数，\(E\)为边集，我们很容易写出状态转移方程

①若\(i \notin S\)

\[f[i][S] = \frac{1}{de[i]}\sum\limits_{(i,j) \in E}(f[j][S] + 1)
\]

②若\(i \in S\)

除非\(\{i\} = S\)，\(f[i][S] = 0\)

否则

\[f[i][S] = \frac{1}{de[i]}\sum\limits_{(i,j) \in E}(f[j][S - \{i\}] + 1)
\]

容易发现转移到的集合\(S'\)要么是\(S\)，要么是更小的集合\(S - \{i\}\)

状态数是\(O(n2^{n})\)，如果我们按\(S\)逐一从小计算，计算当前\(S\)时，转移到的\(S - \{i\}\)则可以直接算出

而如果转移到当前的\(S\)，这个方程则有了后效性

直接高斯消元是\(O(n^{3}2^{n})\)的，我们考虑如hdu Maze那题一样解出式子

在集合\(S\)意义下【为了方便我们就省去S这维】，记\(fa[i]\)为\(i\)的父节点，我们不妨设

\[f[i] = A_if[fa[i]] + B_i
\]

①若\(i \notin S\)

如果\(i\)为叶节点，那么\(A_i = B_i = 1\)

否则有

\[\begin{aligned}
f[i] &= \frac{1}{de[i]}\sum\limits_{(i,j) \in E}(f[j][S] + 1) \\
&= \frac{1}{d[i]}f[fa[i]] + \frac{1}{de[i]}\sum_{i = fa[j]}(A_jf[i] + B_j + 1) \\
&= \frac{1}{d[i] - \sum_{i = fa[j]}A_j}f[fa[i]] + \frac{\sum_{i = fa[j]}(B_j + 1) + 1}{d[i] - \sum_{i = fa[j]}A_j}
\end{aligned}
\]

所以

\[\left\{
\begin{aligned}
A_i &= \frac{1}{d[i] - \sum_{i = fa[j]}A_j} \\
B_i &= \frac{\sum_{i = fa[j]}(B_j + 1) + 1}{d[u] - \sum_{i = fa[j]}A_j}
\end{aligned}
\right.
\]

可以由儿子递推

②若\(i \in S\)

除非\(\{i\} = S\)，此时\(A_i = B_i = 0\)

否则\(A_i = 0\)，\(B_i = \frac{1}{de[i]}\sum\limits_{(i,j) \in E}(f[j][S - \{i\}] + 1)\)

计算出所有\(A_i\)和\(B_i\)后回代可得到\(f[i][S]\)

至此可以\(O(n2^n)\)预处理所有\(f[i][S]\)

然后做到\(O(1)\)回答询问

根本不需要什么minmax容斥，\(O(3^n)\)子集枚举

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 19,maxm = (1 << 19),INF = 1000000000,P = 998244353;
inline int read(){
	int out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
	return out * flag;
}
int h[maxn],ne = 1,de[maxn];
struct EDGE{int to,nxt;}ed[maxn << 1];
inline void build(int u,int v){
	ed[++ne] = (EDGE){v,h[u]}; h[u] = ne;
	ed[++ne] = (EDGE){u,h[v]}; h[v] = ne;
	de[u]++; de[v]++;
}
inline int qpow(int a,int b){
	int re = 1;
	for (; b; b >>= 1,a = 1ll * a * a % P)
		if (b & 1) re = 1ll * re * a % P;
	return re;
}
int in[maxn],f[maxn][maxm],nowS,n,Q,rt,maxv;
int A[maxn],B[maxn],fa[maxn];
void dfs(int u){
	if (de[u] == 1 && u != rt){
		if (in[u]){
			A[u] = 0;
			B[u] = (nowS ^ (1 << u - 1)) ? (f[fa[u]][nowS ^ (1 << u - 1)] + 1) % P : 0;
		}
		else A[u] = 1,B[u] = 1;
		return;
	}
	if (in[u]){
		A[u] = 0;
		if (!(nowS ^ (1 << u - 1))) B[u] = 0;
		else{
			B[u] = 0;
			int dv = qpow(de[u],P - 2),e = (nowS ^ (1 << u - 1));
			Redge(u) B[u] = (B[u] + 1ll * dv * (f[to = ed[k].to][e] + 1) % P) % P;
		}
		Redge(u) if ((to = ed[k].to) != fa[u]){
			fa[to] = u; dfs(to);
		}
		return;
	}
	int sa = 0,sb = 0;
	Redge(u) if ((to = ed[k].to) != fa[u]){
		fa[to] = u; dfs(to);
		sa = (sa + A[to]) % P;
		sb = (sb + B[to] + 1) % P;
	}
	int d = qpow(((de[u] - sa) % P + P) % P,P - 2);
	if (u == rt) A[u] = 0,B[u] = 1ll * sb * d % P;
	else A[u] = d,B[u] = 1ll * (sb + 1) % P * d % P;
}
void dfs2(int u){
	if (u == rt) f[u][nowS] = B[u];
	else f[u][nowS] = (1ll * A[u] * f[fa[u]][nowS] % P + B[u]) % P;
	Redge(u) if ((to = ed[k].to) != fa[u]) dfs2(to);
}
int main(){
	n = read(); Q = read(); rt = read(); maxv = (1 << n) - 1;
	for (int i = 1; i < n; i++) build(read(),read());
	REP(i,n) f[i][0] = 0;
	for (nowS = 1; nowS <= maxv; nowS++){
		REP(i,n) in[i] = ((nowS & (1 << i - 1)) > 0);
		dfs(rt); dfs2(rt);
	}
	while (Q--){
		int k = read(),S = 0;
		while (k--) S |= (1 << (read() - 1));
		printf("%d\n",f[rt][S]);
	}
	return 0;
}