Solution:
由于
\[ x^m = \sum_{i=0}^m{~m~\choose i}{~x~\brace i}i! \]
将所求的式子化成这样，挖掘其性质，考虑是否能从儿子转移(或利用以求得信息)。
\[ \begin{aligned} S(u) &= \sum_{i=1}^ndis(u,i)^k\\ &= \sum_{i=1}^n\sum_{j=0}^k{dis(u, i) \choose j}{k\brace j}j!\\ &= \sum_{j=0}^kj!{k\brace j}\sum_{i=1}^n{dis(u, i)\choose j} \end{aligned} \]
由于组合数有：\({n\choose m} = {n - 1\choose m - 1} + {n - 1\choose m}\)
而从儿子及父亲到自己的距离为１，于是可以考虑换根树型dp求出每个点的 \(\sum_{i=1}^n{dis(u, i)\choose j}\)
设 \(f[u][j] = \sum_{i}{dis(u, i) \choose j}\) 其中 \(i\) 为 \(u\) 子树中的点。
设 \(g[u][j] = \sum_{i=1}^n{dis(u, i)\choose j}\)
\[ f[u][j] = \sum_{v\in son(u)}f[v][j] + f[v][j - 1]\\ g[u][j] = g[fa(u)][j-1]-f[u][j-2]-f[u][j-1]+g[fa(u)][j]-f[u][j-1]-f[u][j]+f[u][j] \]
Code
#include <vector>
#include <cmath>
#include <cstdio>
#include <cassert>
#include <cstring>
#include <iostream>
#include <algorithm>typedef long long LL;
typedef unsigned long long uLL;#define fir first
#define sec second
#define SZ(x) (int)x.size()
#define MP(x, y) std::make_pair(x, y)
#define PB(x) push_back(x)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define GO debug("GO\n")
#define rep(i, a, b) for (register int i = (a), i##end = (b); (i) <= i##end; ++ (i))
#define drep(i, a, b) for (register int i = (a), i##end = (b); (i) >= i##end; -- (i))
#define REP(i, a, b) for (register int i = (a), i##end = (b); (i) < i##end; ++ (i))inline int read() {register int x = 0; register int f = 1; register char c;while (!isdigit(c = getchar())) if (c == '-') f = -1;while (x = (x << 1) + (x << 3) + (c xor 48), isdigit(c = getchar()));return x * f;
}
template<class T> inline void write(T x) {static char stk[30]; static int top = 0;if (x < 0) { x = -x, putchar('-'); }while (stk[++top] = x % 10 xor 48, x /= 10, x);while (putchar(stk[top--]), top);
}
template<typename T> inline bool chkmin(T &a, T b) { return a > b ? a = b, 1 : 0; }
template<typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }using namespace std;const int maxN = 50004;
const int maxK = 153;
const int MOD = 10007;int n, k;
int fac[maxK];
int stirl[maxK][maxK];
vector<int> ver[maxN];void Input()
{n = read(), k = read();for (int i = 1; i < n; ++i){int u = read(), v = read();ver[u].push_back(v);ver[v].push_back(u);}
}void Init()
{fac[0] = 1;rep (i, 1, k) fac[i] = 1ll * fac[i - 1] * i % MOD;stirl[0][0] = stirl[1][1] = 1;rep (i, 2, k)rep (j, 1, i) stirl[i][j] = (1ll * stirl[i - 1][j - 1] + 1ll * j * stirl[i - 1][j] % MOD) % MOD;
}int f[maxN][maxK], g[maxN][maxK], tmp[maxK];void dfs1(int u, int fa)
{f[u][0] = 1;for (int v : ver[u]) if (v != fa){dfs1(v, u);f[u][0] = (1ll * f[u][0] + f[v][0]) % MOD;for (int j = 1; j <= k; ++j)f[u][j] = ((1ll * f[u][j] + f[v][j]) % MOD + f[v][j - 1]) % MOD;}
}void add(int &x, int y)
{x = (1ll * x + y + MOD) % MOD;
}void dfs2(int u, int fa)
{if (!fa) for (int i = 0; i <= k; ++i) g[u][i] = f[u][i];else {g[u][0] = g[fa][0];for (int j = 1; j <= k; ++j){int &x = g[u][j];x = 0;add(x, g[fa][j]);add(x, -f[u][j]);add(x, -f[u][j - 1]);add(x, g[fa][j - 1]);add(x, -f[u][j - 1]);add(x, f[u][j]);if (j >= 2) add(x, -f[u][j - 2]);}}for (int v : ver[u])if (v != fa)dfs2(v, u);
}void Solve()
{dfs1(1, 0);dfs2(1, 0);for (int i = 1; i <= n; ++i){int ans = 0;for (int j = 0; j <= k; ++j)ans = (1ll * ans + 1ll * stirl[k][j] * fac[j] % MOD * g[i][j] % MOD) % MOD;cout << ans << endl;}
}int main()
{
#ifndef ONLINE_JUDGEfreopen("tmp.in", "r", stdin);freopen("tmp.out", "w", stdout);
#endifInput();Init();Solve();return 0;
}
