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P4827 [国家集训队] Crash 的文明世界

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P4827 [国家集训队] Crash 的文明世界

Algorithm I

考虑点分治，每一次考虑跨越当前点的答案，对于一棵子树，对于子树外记录 $cnt _ i$ 表示深度为 $i$ 的点的个数有多少个。那么子树内深度为 $d$ 的答案为 $ans _ d = \sum _ {i} cnt _ i (i + d) ^ k$ ，发现两项差为点值，可以卷积。于是可以在 $O(n \log ^ 2 n)$ 内完成。

查看代码 
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#include <cstdio>
#include <vector>
#define pb push_back
using namespace std;
template <class Type>
void read (Type &x)
{
    char c;
    bool flag = false;
    while ((c = getchar()) < '0' || c > '9')
        c == '-' && (flag = true);
    x = c - '0';
    while ((c = getchar()) >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + c - '0';
    flag && (x = ~x + 1);
}
template <class Type, class ...rest>
void read (Type &x, rest &...y) { read(x), read(y...); }
template <class Type>
void write (Type x)
{
    x < 0 && (putchar('-'), x = ~x + 1);
    x > 9 && (write(x / 10), 0);
    putchar('0' + x % 10);
}
typedef long long LL;
const int N = 5e4 + 10, mod = 1e4 + 7;
namespace Mul
{
    const int N = 2e5 + 10, mod = 998244353;
    int rev[N];
    int binpow (int b, int k = mod - 2)
    {
        int res = 1;
        for (; k; k >>= 1, b = (LL)b * b % mod)
            if (k & 1) res = (LL)res * b % mod;
        return res;
    }
    void ntt (int *x, int bit, int op)
    {
        int tot = 1 << bit;
        for (int i = 0; i < tot; ++i)
            if ((rev[i] = rev[i >> 1] >> 1 | (i & 1) << bit - 1) > i)
                swap(x[i], x[rev[i]]);
        for (int mid = 1; mid < tot; mid <<= 1)
        {
            int w1 = binpow(3, (mod - 1) / (mid << 1));
            if (!~op) w1 = binpow(w1);
            for (int i = 0; i < tot; i += mid << 1)
                for (int j = 0, k = 1; j < mid; ++j, k = (LL)k * w1 % mod)
                {
                    int p = x[i | j], q = (LL)k * x[i | mid | j] % mod;
                    x[i | j] = (p + q) % mod, x[i | mid | j] = (p - q) % mod;
                }
        }
        if (~op) return;
        int itot = binpow(tot);
        for (int i = 0; i < tot; ++i)
            x[i] = (LL)x[i] * itot % mod;
    }
    void PolyMerge (int n, int *f, int *g, int *w)
    {
        int bit = 0;
        while (1 << bit < n + n - 1) ++bit;
        int tot = 1 << bit;
        static int A[N], B[N], C[N];
        for (int i = 0; i < n; ++i) A[i] = f[n - i - 1];
        for (int i = n; i < tot; ++i) A[i] = 0;
        for (int i = 0; i < n; ++i) B[i] = g[i];
        for (int i = n; i < tot; ++i) B[i] = 0;
        ntt(A, bit, 1), ntt(B, bit, 1);
        for (int i = 0; i < tot; ++i)
            C[i] = (LL)A[i] * B[i] % mod;
        ntt(C, bit, -1);
        for (int i = 0; i < n; ++i)
            w[i] = (C[n - i - 1] + mod) % mod;
    }
}
bool vis[N];
vector <int> g[N];
int n, m, ans[N], v[N], h[N], cnt[N];
int binpow (int b, int k)
{
    int res = 1;
    for (; k; k >>= 1, b = (LL)b * b % mod)
        if (k & 1) res = (LL)res * b % mod;
    return res;
}
int WeightCentre (int x, int fa, int tot, int &wc)
{
    if (vis[x]) return 0;
    int sz = 1, mx = 0;
    for (int i : g[x]) if (i ^ fa)
    {
        int t = WeightCentre(i, x, tot, wc);
        mx = max(mx, t), sz += t;
    }
    mx = max(mx, tot - sz);
    if (mx <= tot / 2) wc = x;
    return sz;
}
int Size (int x, int fa)
{
    if (vis[x]) return 0;
    int sz = 1;
    for (int i : g[x]) if (i ^ fa)
        sz += Size(i, x);
    return sz;
}
void add (int x, int fa, int d, int &mx)
{
    if (vis[x]) return;
    mx = max(mx, d);
    ++cnt[d++];
    for (int i : g[x]) if (i ^ fa)
        add(i, x, d, mx);
}
void del (int x, int fa, int d)
{
    if (vis[x]) return;
    --cnt[d++];
    for (int i : g[x]) if (i ^ fa)
        del(i, x, d);
}
void get (int x, int fa, int d)
{
    if (vis[x]) return;
    ans[x] = (ans[x] + h[d]) % mod;
    ++cnt[d++];
    for (int i : g[x]) if (i ^ fa)
        get(i, x, d);
}
void calc (int x)
{
    if (vis[x]) return;
    WeightCentre(x, 0, Size(x, 0), x);
    int d = 0;
    add(x, 0, 0, d);
    vis[x] = true;
    for (int i = 1; i <= d; ++i)
        ans[x] = (ans[x] + cnt[i] * v[i]) % mod;
    for (int i : g[x])
    {
        del(i, x, 1);
        Mul::PolyMerge(d * 2 + 1, v, cnt, h);
        get(i, x, 1);
    }
    for (int i = 0; i <= d; ++i) cnt[i] = 0;
    for (int i : g[x]) calc(i);
}
int main ()
{
    read(n, m);
    for (int i = 1; i <= n; ++i)
        v[i] = binpow(i, m);
    for (int i = 1, a, b; i < n; ++i)
        read(a, b), g[a].pb(b), g[b].pb(a);
    calc(1);
    for (int i = 1; i <= n; ++i)
        write(ans[i]), puts("");
    return 0;
}

Algorithm II

注意到幂的形式以及 $k$ 并不大，考虑将其拆为斯特林数和下降幂。

$$
\begin {aligned}
S(x) & = \sum _ {j = 1} ^ n d(x, j) ^ k \\
& = \sum _ {j = 1} ^ n \sum _ {i = 0} ^ k {k \brace i} d(x, j) ^ {\underline i} \\
& = \sum _ {i = 0} ^ k i! {k \brace i} \sum _ {j = 1} ^ n \binom {d(x, j)} i
\end {aligned}
$$

考虑 $\binom n m $ 的递推式，$\binom n m = \binom {n - 1} {m - 1} + \binom {n - 1} m$ 。考虑每个点为根的情况，结合 $d(x, j)$ 的意义，记 $f _ {i, j}$ 表示 $i$ 子树内，组合数下标为 $j$ 的和，那么有 $f _ {i, j} = \sum _ {k \in son _ i} f _ {i, j} + f _ {i, j - 1} $ 。得到 $1$ 为根的答案后，换根DP即可。

查看代码 
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#include <cstdio>
#include <vector>
#define pb push_back
using namespace std;
template <class Type>
void read (Type &x)
{
    char c;
    bool flag = false;
    while ((c = getchar()) < '0' || c > '9')
        c == '-' && (flag = true);
    x = c - '0';
    while ((c = getchar()) >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + c - '0';
    flag && (x = ~x + 1);
}
template <class Type, class ...rest>
void read (Type &x, rest &...y) { read(x), read(y...); }
template <class Type>
void write (Type x)
{
    x < 0 && (putchar('-'), x = ~x + 1);
    x > 9 && (write(x / 10), 0);
    putchar('0' + x % 10);
}
const int N = 5e4 + 10, M = 160, mod = 1e4 + 7;
int n, m, f[N][M], g[N][M], s[M][M], fact[M];
vector <int> e[N];
void init ()
{
    s[0][0] = 1;
    for (int i = 1; i <= m; ++i)
        for (int j = 1; j <= m; ++j)
            s[i][j] = (s[i - 1][j - 1] + j * s[i - 1][j]) % mod;
    fact[0] = 1;
    for (int i = 1; i <= m; ++i)
        fact[i] = fact[i - 1] * i % mod;
}
void dfs1 (int u = 1, int fa = 0)
{
    f[u][0] = 1;
    for (int v : e[u]) if (v ^ fa)
    {
        dfs1(v, u);
        (f[u][0] += f[v][0]) %= mod;
        for (int i = 1; i <= m; ++i)
            (f[u][i] += f[v][i] + f[v][i - 1]) %= mod;
    }
}
void dfs2 (int u = 1, int fa = 0)
{
    for (int v : e[u]) if (v ^ fa)
    {
        static int t[N];
        t[0] = (g[u][0] - f[v][0]);
        for (int i = 1; i <= m; ++i)
            t[i] = (g[u][i] - f[v][i] - f[v][i - 1]) % mod;
        g[v][0] = (f[v][0] + t[0]);
        for (int i = 1; i <= m; ++i)
            g[v][i] = (f[v][i] + t[i] + t[i - 1]) % mod;
        dfs2(v, u);
    }
}
int main ()
{
    read(n, m);
    init();
    for (int i = 1, a, b; i < n; ++i)
        read(a, b), e[a].pb(b), e[b].pb(a);
    dfs1();
    for (int i = 0; i <= m; ++i) g[1][i] = f[1][i];
    dfs2();
    for (int i = 1; i <= n; ++i)
    {
        int res = 0;
        for (int j = 0; j <= m; ++j)
            res = (res + s[m][j] * fact[j] % mod * g[i][j]) % mod;
        write((res + mod) % mod), puts("");
    }
    return 0;
}
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