loj#2542. 「PKUWC2018」随机游走(MinMax容斥 期望dp)

题意

题目链接

Sol

考虑直接对询问的集合做MinMax容斥

设\(f[i][sta]\)表示从\(i\)到集合\(sta\)中任意一点的最小期望步数

按照树上高斯消元的套路，我们可以把转移写成\(f[x] = a_x f[fa] + b_x\)的形式

然后直接推就可以了

更详细的题解

#include<bits/stdc++.h> 
#define LL long long 
using namespace std;
const int MAXN = 1e6 + 10, mod = 998244353;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int mul(int x, int y) {return 1ll * x * y % mod;}
int add(int x, int y) {if(x + y < 0) return x + y + mod; else return x + y >= mod ? x + y - mod : x + y;}
int fp(int a, int p) {
    int base = 1;
    for(; p; p >>= 1, a = mul(a, a))
        if(p & 1) base = mul(base, a);
    return base;
}
int inv(int x) {
    x = (x + mod) % mod;
    return fp(x, mod - 2);
}
int N, Q, S, Lim, g[MAXN], deg[MAXN], a[MAXN], b[MAXN], siz[MAXN];
vector<int> v[MAXN];
void dfs(int x, int fa, int sta) {
    if(sta & (1 << x - 1)) {a[x] = b[x] = 0; return ;}
    int ta = 0, tb = 0;
    for(int i = 0; i < v[x].size(); i++) {
        int to = v[x][i]; if(to == fa) continue;
        dfs(to, x, sta);
        ta += a[to]; tb += b[to];
    }
    a[x] = inv(deg[x] - ta); 
    b[x] = mul((tb + deg[x]), inv(deg[x] - ta));
}
int main() {
    N = read(); Q = read(); S = read(); Lim = (1 << N) - 1;
    for(int i = 1; i <= N - 1; i++) {
        int x = read(), y = read();
        v[x].push_back(y); v[y].push_back(x);
        deg[x]++; deg[y]++;
    }
    for(int sta = 1; sta <= Lim; sta++) {
        siz[sta] = siz[sta >> 1] + (sta & 1);
        dfs(S, 0, sta);
        g[sta] = b[S];
    }
    while(Q--) {
        int k = read(), S = 0, ans = 0;
        for(int i = 1; i <= k; i++) S |= (1 << (read() - 1));
        for(int i = S; i; i = (i - 1) & S) {
            if(siz[i] & 1) ans = add(ans, g[i]);
            else ans = add(ans, -g[i]);
        }
        printf("%d\n", ans);
    }
    return 0;
}