LOJ#2542 随机游走

解：首先minmax容斥变成经过集合t的第一个点就停止的期望步数。对于某个t，设从x开始的期望步数为f(x)

如果x∈t，f(x) = 0。否则f(x) = ∑f(y) / in[x] + 1

树上高斯消元。从叶子往上，可以发现每个点都可以表示为Af(fa) + B

于是我们推一波式子，参考，就可以对每个t，O(n)求出f(root)。

然后每个询问就枚举子集。

注意DFS的时候可以剪枝，遇到x∈t就返回，否则T飞.....

#include <bits/stdc++.h>

const int N = 30, MO = 998244353;

inline void read(int &x) {
    x = 0;
    char c = getchar();
    while(c < '0' || c > '9') c = getchar();
    while(c >= '0' && c <= '9') {
        x = x * 10 + c - 48;
        c = getchar();
    }
    return;
}

struct Edge {
    int nex, v;
}edge[N << 1]; int tp;

int n, rt, e[N], A[N], B[N], now, in[N], ans[1 << 19], cnt[1 << 19], ans2[1 << 19];

inline int qpow(int a, int b) {
    int ans = 1;
    a = (a + MO) % MO;
    while(b) {
        if(b & 1) ans = 1ll * ans * a % MO;
        a = 1ll * a * a % MO;
        b = b >> 1;
    }
    return ans;
}

inline void add(int x, int y) {
    tp++;
    edge[tp].v = y;
    edge[tp].nex = e[x];
    e[x] = tp;
    return;
}

void DFS(int x, int f) {
    if(((1 << (x - 1)) | now) == now) {
        A[x] = B[x] = 0;
        return;
    }
    int sa = 0, sb = 0;
    for(int i = e[x]; i; i = edge[i].nex) {
        int y = edge[i].v;
        if(y == f) continue;
        DFS(y, x);
        sa = (sa + A[y]) % MO;
        sb = (sb + B[y]) % MO;
    }

    A[x] = qpow(in[x] - sa, MO - 2);
    B[x] = 1ll * A[x] * (in[x] + sb) % MO;

    return;
}

int main() {
    int q;
    read(n); read(q); read(rt);
    for(register int i = 1, x, y; i < n; i++) {
        read(x); read(y);
        add(x, y); add(y, x);
        in[x]++; in[y]++;
    }

    int lm = 1 << n;
    /*for(now = 1; now < lm; now++) {
        //memset(A, )
        DFS(rt, 0);
        ans[now] = B[rt];
    }*/
    memset(ans, -1, sizeof(ans));
    memset(ans2, -1, sizeof(ans2));

    for(register int i = 1; i < lm; i++) {
        cnt[i] = 1 + cnt[i - (i & (-i))];
    }

    /// prework OVER
    for(register int i = 1, k; i <= q; i++) {
        read(k);
        int s = 0;
        for(register int j = 1, x; j <= k; j++) {
            read(x);
            s |= (1 << (x - 1));
        }
        int Ans = 0;
        if(ans2[s] != -1) {
            Ans = ans2[s];
        }
        else {
            for(register int t = s; t; t = (t - 1) & s) {
                if(ans[t] == -1) {
                    now = t;
                    DFS(rt, 0);
                    ans[t] = B[rt];
                }
                if(cnt[t] & 1) Ans = (Ans + ans[t]) % MO;
                else Ans = (Ans - ans[t] + MO) % MO;
            }
            ans2[s] = Ans;
        }
        printf("%d
", (Ans + MO) % MO);
    }

    return 0;
}