SP7363 TREESUM - Tree Sum 题解

$$\begin{array}{rl}& \sum _{i\in \text{subtree(u)}}d(u,i{)}^k\\ =& \sum _{i\in \text{subtree(u)}}\sum _{j=0}^k(\genfrac{}{}{0ex}{}{d(u,i)}{j})\left\{\begin{array}{c}k\\ j\end{array}\right\}j!\\ =& \sum _{j=0}^k\left\{\begin{array}{c}k\\ j\end{array}\right\}j!\sum _{i\in \text{subtree(u)}}(\genfrac{}{}{0ex}{}{d(u,i)}{j})\end{array}$$

设 $f_{u,j}=\sum _{i\in \text{subtree(u)}}(\genfrac{}{}{0ex}{}{d(u,i)}{j})$，则 $f_{u,j}=\sum _{i\in \text{subtree(u)}}(\genfrac{}{}{0ex}{}{d(u,i)}{j})=\sum _{i\in \text{subtree(u)}}(\genfrac{}{}{0ex}{}{d(u,i)-1}{j})-\sum _{i\in \text{subtree(u)}}(\genfrac{}{}{0ex}{}{d(u,i)-1}{j-1})=f_{v,j}-f_{v,j-1}$。

然后这是子树内的，换根一遍得到全局的。

#include <cstdio>
#include <cstring>
#define M 1000000007
#define int long long
struct E
{
    int v, t;
} e[40050];
int T, n, K, c, q, o[250], s[250][250], h[20050], f[20050][250], g[20050][250];
void A(int u, int v)
{
    e[++c] = {v, h[u]};
    h[u] = c;
}
void F(int u, int k)
{
    f[u][0] = f[u][1] = 1;
    for (int i = h[u], v; i; i = e[i].t)
        if ((v = e[i].v) != k)
        {
            F(v, u);
            f[u][0] += f[v][0];
            for (int j = 1; j <= K; ++j)
                f[u][j] = (f[u][j] + f[v][j] + f[v][j - 1]) % M;
        }
}
void G(int u, int k)
{
    for (int i = h[u], v; i; i = e[i].t)
        if ((v = e[i].v) != k)
        {
            g[u][0] -= f[v][0];
            for (int j = 1; j <= K; ++j)
                g[u][j] = (g[u][j] + M - f[v][j] + M - f[v][j - 1]) % M;
            for (int j = 0; j <= K; ++j)
                g[v][j] = f[v][j];
            g[v][0] += g[u][0];
            for (int j = 1; j <= K; ++j)
                g[v][j] = (g[v][j] + g[u][j] + g[u][j - 1]) % M;
            g[u][0] += f[v][0];
            for (int j = 1; j <= K; ++j)
                g[u][j] = (g[u][j] + f[v][j] + f[v][j - 1]) % M;
            G(v, u);
        }
}
signed main()
{
    scanf("%lld", &T);
    while (T--)
    {
        c = 0;
        memset(h, 0, sizeof h);
        memset(f, 0, sizeof f);
        memset(g, 0, sizeof g);
        scanf("%lld%lld", &n, &K);
        for (int i = 1, u, v; i < n; ++i)
            scanf("%lld%lld", &u, &v), ++u, ++v, A(u, v), A(v, u);
        F(1, 0);
        for (int i = o[0] = 1; i <= K; ++i)
            o[i] = o[i - 1] * i % M;
        s[0][0] = s[1][1] = 1;
        for (int i = 2; i <= K; ++i)
            for (int j = 1; j <= i; ++j)
                s[i][j] = (s[i - 1][j - 1] + j * s[i - 1][j]) % M;
        q = 0;
        for (int i = 0; i <= K; ++i)
            g[1][i] = f[1][i];
        G(1, 0);
        for (int i = 1; i <= n; ++i, printf("%lld\n", q))
            for (int j = q = 0; j <= K; ++j)
                q = (q + s[K][j] * o[j] % M * g[i][j]) % M;
        puts("");
    }
    return 0;
}