LOJ2542. 「PKUWC2018」随机游走【概率期望DP+Min-Max容斥（最值反演）】

题面

---

思路

我们可以把到每个点的期望步数算出来取max？但是直接算显然是不行的

那就可以用Min-Max来容斥一下

设\(g_{s}\)是从x到s中任意一个点的最小步数

设\(f_{s}\)是从x到s中任意一个点的最大步数

然后就可以的得到

\(f_{s}=\sum_{t\subseteq s}(-1)^{|t|+1}g_t\)

然后考虑g怎么求

设\(p_i\)是i点到任意一个子集中的点的最小步数

有\(p_u=\frac{1}{du_u}(1+p_{fa_u})+\frac{1}{du_u}\sum_{v\in child_u}(p_v+1)\)

然后我们令\(p_u=a_up_{fa_u}+b_u\)

很显然有\(p_u=\frac{1}{du_u}\sum(a_vf_u+b_v+1)+\frac{1}{du_u}(p_{fa_u})\)

然后移项可以得到\(a_u=\frac{1}{du_u-\sum a_v},b_u=\frac{\sum(b_v+1)+1}{du_u-\sum a_v}\)

然后因为x是根没有父亲，所以\(g_{s}=(bitcnt(s) \& 1)?b_u:-b_u\)

然后就可以用子集前缀和进行累加了

最后直接输出答案就可以了

---

#include <bits/stdc++.h>

using namespace std;

const int Mod = 998244353;
const int N = 20;

int n, m, x;
int a[N], b[N], du[N];
int f[1 << N];
vector<int> g[N];

int main() {
#ifdef dream_maker
  freopen("input.txt", "r", stdin);
#endif
  function<int(int a, int b)> add = [&](int a, int b) {
    return (a += b) >= Mod ? a - Mod : a;
  };
  
  function<int(int a, int b)> sub = [&](int a, int b) {
    return (a -= b) < 0 ? a + Mod : a;
  };
  
  function<int(int a, int b)> mul = [&](int a, int b) {
    return (long long) a * b % Mod;
  };
  
  function<int(int a, int b)> fast_pow = [&](int a, int b) {
    int res = 1;
    for (; b; b >>= 1, a = mul(a, a))
      if (b & 1) res = mul(res, a);
    return res;
  };
  
  function<int(int a)> bitcnt = [&](int a) {
    int res = 0;
    for (; a; a >>= 1)
      if (a & 1) ++res;
    return res;
  };
  
  function<void(int u, int fa, int s)> dfs = [&](int u, int fa, int s) {
    if ((s >> (u - 1)) & 1) return;
    a[u] = du[u], b[u] = (u == x) ? 0 : 1; // x不用向fa走的1 
    for (auto v : g[u]) {
      if (v == fa) continue;
      dfs(v, u, s);
      a[u] = sub(a[u], a[v]);
      b[u] = add(b[u], b[v] + 1);
    }
    a[u] = fast_pow(a[u], Mod - 2);
    b[u] = mul(b[u], a[u]);
  };
  
  scanf("%d %d %d", &n, &m, &x);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
    ++du[u], ++du[v];
  }
  int up = (1 << n) - 1;
  for (int s = 1; s <= up; s++) {
    for (int i = 1; i <= n; i++)
      a[i] = b[i] = 0;
    dfs(x, 0, s);
    f[s] = (bitcnt(s) & 1) ? b[x] : (Mod - b[x]) % Mod;
  }
  f[0] = 0;
  for (int i = 1; i <= n; i++) { // 这个循环在外面 
    for (int s = 1; s <= up; s++) {
      if ((s >> (i - 1)) & 1) {
        f[s] = add(f[s], f[s ^ (1 << (i - 1))]);
      }
    }
  }
  while (m--) {
    int num, cur, s = 0;
    scanf("%d", &num);
    while (num--) {
      scanf("%d", &cur);
      s |= 1 << (cur - 1);
    }
    printf("%d\n", f[s]);
  }
  return 0;
}