[HDU 4625]JZPTREE

danihao123 posted @ 2018年3月24日 22:55 in 题解 with tags HDU 树形DP 第二类斯特林数 , 606 阅读
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第一次用到第二类斯特林数……

首先众所周知第二类斯特林数有一个很好康的式子:

\[x^k = \sum_{i = 0}^k S(k, i) x^{\underline{i}}\]

然后带进去试试?一番化简之后得到:

\[E_u = \sum_{i = 0}^k S(k, i) \sum_{v = 1}^n dist(u, v)^{\underline{i}}\]

那个下降幂很恶心……把它强行搞成组合数后发现:

\[E_u = \sum_{i = 0}^k S(k, i) i! \sum_{v = 1}^n \binom{dist(u, v)}{i}\]

然后考虑每个点预处理后面那个和式……显然可以树形DP苟。用经典的树形DP套路,设两个状态分别表示一个点的子树内部的贡献以及外面的点给他的贡献。至于转移,由于那是个组合数,所以可以考虑用Pascal定理苟……

代码:

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cctype>
#include <cmath>
#include <algorithm>
#include <utility>
const int maxn = 50005;
const int maxk = 505;
const int ha = 10007;
int first[maxn];
int next[maxn << 1], to[maxn << 1];
int cnt;
void add_edge(int u, int v) {
  cnt ++;
  next[cnt] = first[u]; first[u] = cnt;
  to[cnt] = v;
}

int n, k;
int down[maxn][maxk];
void dfs_1(int x, int fa = 0) {
  down[x][0] = 1;
  for(int i = first[x]; i; i = next[i]) {
    int v = to[i];
    if(v != fa) {
      dfs_1(v, x);
      down[x][0] = (down[x][0] + down[v][0]) % ha;
      for(int j = 1; j <= k; j ++) {
        down[x][j] = (down[x][j] + down[v][j] + down[v][j - 1]) % ha;
      }
    }
  }
}
int up[maxn][maxk];
void dfs_2(int x, int fa = 0) {
  if(fa) {
    up[x][0] = n - down[x][0];
    for(int j = 1; j <= k; j ++) {
      up[x][j] = (up[fa][j] + up[fa][j - 1] + down[fa][j] + down[fa][j - 1]) % ha;
      up[x][j] = (up[x][j] - (2 * down[x][j - 1]) % ha + ha) % ha;
      up[x][j] = (up[x][j] - down[x][j] + ha) % ha;
      if(j > 1) up[x][j] = (up[x][j] - down[x][j - 2] + ha) % ha;
    }
  }
  for(int i = first[x]; i; i = next[i]) {
    int v = to[i];
    if(v != fa) {
      dfs_2(v, x);
    }
  }
}

int S[maxk][maxk];
int fac[maxn];
void process() {
  S[1][1] = 1;
  for(int i = 2; i <= k; i ++) {
    S[i][0] = 0;
    for(int j = 1; j < i; j ++) {
      S[i][j] = ((j * S[i - 1][j]) % ha + S[i - 1][j - 1]) % ha;
    }
    S[i][i] = 1;
  }
  fac[0] = 1;
  for(int i = 1; i <= n; i ++) {
    fac[i] = (fac[i - 1] * (i % ha)) % ha;
  }
}

int main() {
  int T; scanf("%d", &T);
  n = 50000; k = 500; process();
  while(T --) {
    scanf("%d%d", &n, &k);
    cnt = 0; memset(first, 0, sizeof(first));
    for(int i = 0; i < n - 1; i ++) {
      int u, v; scanf("%d%d", &u, &v);
      add_edge(u, v); add_edge(v, u);
    }
    memset(down, 0, sizeof(down));
    memset(up, 0, sizeof(up));
    dfs_1(1); dfs_2(1);
    for(int i = 1; i <= n; i ++) {
      int ans = 0;
      for(int j = 1; j <= k; j ++) {
        int f_v = (down[i][j] + up[i][j]) % ha;
        ans = (ans + (S[k][j] * ((f_v * fac[j]) % ha)) % ha) % ha;
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}
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Sep 12, 2022 05:28:14 PM

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