题目简述:给定$n \leq 50000$个节点的数,每条边的长度为$1$,对每个节点$u$,求

$$ E_u = \sum_{v=1}^n (d(u, v))^k, $$

其中$d(u, v)$是节点$u$和节点$v$的距离,而$k \leq 500$是一个常数。

解1:

斯特林数的性质,我们注意到

$$ x^n = \sum_{k=0}^n \begin{Bmatrix} n \\ k \end{Bmatrix} x^{\underline{k}}. $$

从而,

$$ E_u = \sum_{v=1}^n (d(u, v))^k = \sum_{i=0}^k \begin{Bmatrix} k \\ i \end{Bmatrix} \sum_{v=1}^n (d(u, v))^{\underline{i}}. $$

为此,我们定义

$$f[u][k] = \sum_{v \in T_u} (d(u, v))^{\underline{k}},$$

其中$T_u$表示以$u$为根节点的子树。令$\text{son}(u)$表示节点$u$的所有儿子节点的集合,并注意到$(x+1)^{\underline{k}} = x^{\underline{k}}+kx^{\underline{k-1}}$,则

$$
\begin{aligned}
f[u][k]
& = \sum_{v \in \text{son}(u)} \sum_{w \in T_v} (d(u, w))^{\underline{k}} \\
& = \sum_{v \in \text{son}(u)} \sum_{w \in T_v} (d(v, w)+1)^{\underline{k}} \\
& = \sum_{v \in \text{son}(u)} \sum_{w \in T_v} \Big( (d(v, w))^{\underline{k}}+k (d(v, w))^{\underline{k-1}} \Big) \\
& = \sum_{v \in \text{son}(u)} \Big( f[v][k]+k f[v][k-1] \Big)
\end{aligned}
$$

两遍DFS即可求出所有$E_u$,从而可在$O(nk)$的复杂度内解决。

 #include <bits/stdc++.h>

 using namespace std;

 typedef long long ll;
typedef unsigned long long ull;
typedef double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<ld,ld> pdd; #define X first
#define Y second //#include <boost/unordered_map.hpp>
//using namespace boost; /*
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> rbtree;
rbtree T;
*/ namespace io{
const int L = ( << ) + ; char buf[L], *S , *T, c; char getchar() {
if(__builtin_expect(S == T, )) {
T = (S = buf) + fread(buf, , L, stdin);
return (S == T ? EOF : *S++);
}
return *S++;
} int inp() {
int x = , f = ; char ch;
for(ch = getchar(); !isdigit(ch); ch = getchar())
if(ch == '-') f = -;
for(; isdigit(ch); x = x * + ch - '', ch = getchar());
return x * f;
} unsigned inpu()
{
unsigned x = ; char ch;
for(ch = getchar(); !isdigit(ch); ch = getchar());
for(; isdigit(ch); x = x * + ch - '', ch = getchar());
return x;
} ll inp_ll() {
ll x = ; int f = ; char ch;
for(ch = getchar(); !isdigit(ch); ch = getchar())
if(ch == '-') f = -;
for(; isdigit(ch); x = x * + ch - '', ch = getchar());
return x * f;
} char B[], *outs=B+, *outr=B+;
template<class T>
inline void print(register T a,register char x=){
if(x) *--outs = x, x = ; if(!a)*--outs = '';
else
while(a)
*--outs = (a % ) + , a /= ; if(x)
*--outs = x; fwrite(outs, outr - outs , , stdout);
outs = outr;
}
}; using io :: print;
using io :: inp;
using io :: inpu;
using io :: inp_ll; using i32 = int;
using i64 = long long;
using u8 = unsigned char;
using u32 = unsigned;
using u64 = unsigned long long;
using f64 = double;
using f80 = long double; ll power(ll a, ll b, ll p)
{
if (!b) return ;
ll t = power(a, b/, p);
t = t*t%p;
if (b&) t = t*a%p;
return t;
} ll exgcd(ll a, ll b, ll &x, ll &y)
{
if (b == )
{
x = ;
y = ;
return a;
}
ll px, py;
ll d = exgcd(b, a%b, px, py);
x = py;
y = px-a/b*py;
return d;
} template<class T>
inline void freshmin(T &a, const T &b)
{
if (a > b) a = b;
} template<class T>
inline void freshmax(T &a, const T &b)
{
if (a < b) a = b;
} const int MAXN = ;
const int MAXK = ;
const int MOD = ;
const f80 MI = f80()/MOD;
const int INF = ; int n, k;
int S[MAXK][MAXK]; vector<int> v[MAXN];
int f[MAXN][MAXK], g[MAXN][MAXK]; void dfs1(int x, int p)
{
f[x][] = ;
for (int i = ; i <= k; ++ i)
f[x][i] = ;
for (auto y : v[x])
{
if (y == p) continue;
dfs1(y, x);
(f[x][] += f[y][]) %= MOD;
for (int i = ; i <= k; ++ i)
(f[x][i] += f[y][i]+i*f[y][i-]) %= MOD;
}
} void dfs2(int x, int p)
{
if (!p)
{
for (int i = ; i <= k; ++ i)
g[x][i] = f[x][i];
}
for (auto y : v[x])
{
if (y == p) continue;
g[y][] = g[x][];
for (int i = ; i <= k; ++ i)
{
int g1 = (g[x][i]-(f[y][i]+i*f[y][i-]))%MOD;
int g2 = (g[x][i-]-(f[y][i-]+(i-)*(i- >= ? f[y][i-] : )))%MOD;
g[y][i] = (f[y][i]+g1+i*g2)%MOD;
}
dfs2(y, x);
}
} int main()
{ S[][] = ;
for (int i = ; i <= ; ++ i)
for (int j = ; j <= i; ++ j)
S[i][j] = (S[i-][j-]+S[i-][j]*j)%MOD; for (int T = inp(); T --; )
{
n = inp();
k = inp();
for (int i = ; i <= n; ++ i)
v[i].clear();
for (int i = ; i < n; ++ i)
{
int x = inp();
int y = inp();
v[x].push_back(y);
v[y].push_back(x);
}
dfs1(, );
dfs2(, );
for (int x = ; x <= n; ++ x)
{
int ret = ;
for (int i = ; i <= k; ++ i)
(ret += S[k][i]*g[x][i]) %= MOD;
printf("%d\n", (ret+MOD)%MOD);
}
} return ;
}

解2:

我们用另一个斯特林数的性质:

$$ x^n = \sum_{k=0}^n k! \begin{Bmatrix} n \\ k \end{Bmatrix} \binom{x}{k}. $$

从而,

$$ E_u = \sum_{v=1}^n (d(u, v))^k = \sum_{i=0}^k i! \begin{Bmatrix} k \\ i \end{Bmatrix} \sum_{v=1}^n \binom{d(u, v)}{i}. $$

为此,我们定义

$$f[u][k] = \sum_{v \in T_u} \binom{d(u, v)}{k},$$

其中$T_u$表示以$u$为根节点的子树。令$\text{son}(u)$表示节点$u$的所有儿子节点的集合,则

$$
\begin{aligned}
f[u][k] 
& = \sum_{v \in \text{son}(u)} \sum_{w \in T_v} \binom{d(u, w)}{k} \\
& = \sum_{v \in \text{son}(u)} \sum_{w \in T_v} \binom{d(v, w)+1}{k} \\
& = \sum_{v \in \text{son}(u)} \sum_{w \in T_v} \left( \binom{d(v, w)}{k} + \binom{d(v, w)}{k-1} \right) \\
& = \sum_{v \in \text{son}(u)} \Big( f[v][k]+f[v][k-1] \Big)
\end{aligned}
$$

两遍DFS即可求出所有$E_u$,从而可在$O(nk)$的复杂度内解决。

 #include <bits/stdc++.h>

 using namespace std;

 typedef long long ll;
typedef unsigned long long ull;
typedef double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<ld,ld> pdd; #define X first
#define Y second //#include <boost/unordered_map.hpp>
//using namespace boost; /*
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> rbtree;
rbtree T;
*/ namespace io{
const int L = ( << ) + ; char buf[L], *S , *T, c; char getchar() {
if(__builtin_expect(S == T, )) {
T = (S = buf) + fread(buf, , L, stdin);
return (S == T ? EOF : *S++);
}
return *S++;
} int inp() {
int x = , f = ; char ch;
for(ch = getchar(); !isdigit(ch); ch = getchar())
if(ch == '-') f = -;
for(; isdigit(ch); x = x * + ch - '', ch = getchar());
return x * f;
} unsigned inpu()
{
unsigned x = ; char ch;
for(ch = getchar(); !isdigit(ch); ch = getchar());
for(; isdigit(ch); x = x * + ch - '', ch = getchar());
return x;
} ll inp_ll() {
ll x = ; int f = ; char ch;
for(ch = getchar(); !isdigit(ch); ch = getchar())
if(ch == '-') f = -;
for(; isdigit(ch); x = x * + ch - '', ch = getchar());
return x * f;
} char B[], *outs=B+, *outr=B+;
template<class T>
inline void print(register T a,register char x=){
if(x) *--outs = x, x = ; if(!a)*--outs = '';
else
while(a)
*--outs = (a % ) + , a /= ; if(x)
*--outs = x; fwrite(outs, outr - outs , , stdout);
outs = outr;
}
}; using io :: print;
using io :: inp;
using io :: inpu;
using io :: inp_ll; using i32 = int;
using i64 = long long;
using u8 = unsigned char;
using u32 = unsigned;
using u64 = unsigned long long;
using f64 = double;
using f80 = long double; ll power(ll a, ll b, ll p)
{
if (!b) return ;
ll t = power(a, b/, p);
t = t*t%p;
if (b&) t = t*a%p;
return t;
} ll exgcd(ll a, ll b, ll &x, ll &y)
{
if (b == )
{
x = ;
y = ;
return a;
}
ll px, py;
ll d = exgcd(b, a%b, px, py);
x = py;
y = px-a/b*py;
return d;
} template<class T>
inline void freshmin(T &a, const T &b)
{
if (a > b) a = b;
} template<class T>
inline void freshmax(T &a, const T &b)
{
if (a < b) a = b;
} const int MAXN = ;
const int MAXK = ;
const int MOD = ;
const f80 MI = f80()/MOD;
const int INF = ; int n, k;
int S[MAXK][MAXK]; vector<int> v[MAXN];
int f[MAXN][MAXK], g[MAXN][MAXK]; void dfs1(int x, int p)
{
f[x][] = ;
for (int i = ; i <= k; ++ i)
f[x][i] = ;
for (auto y : v[x])
{
if (y == p) continue;
dfs1(y, x);
(f[x][] += f[y][]) %= MOD;
for (int i = ; i <= k; ++ i)
(f[x][i] += f[y][i]+f[y][i-]) %= MOD;
}
} void dfs2(int x, int p)
{
if (!p)
{
for (int i = ; i <= k; ++ i)
g[x][i] = f[x][i];
}
for (auto y : v[x])
{
if (y == p) continue;
g[y][] = g[x][];
for (int i = ; i <= k; ++ i)
{
int g1 = (g[x][i]-(f[y][i]+f[y][i-]))%MOD;
int g2 = (g[x][i-]-(f[y][i-]+(i- >= ? f[y][i-] : )))%MOD;
g[y][i] = (f[y][i]+g1+g2)%MOD;
}
dfs2(y, x);
}
} int main()
{ S[][] = ;
for (int i = ; i <= ; ++ i)
for (int j = ; j <= i; ++ j)
S[i][j] = (S[i-][j-]+S[i-][j]*j)%MOD; for (int T = inp(); T --; )
{
n = inp();
k = inp();
for (int i = ; i <= n; ++ i)
v[i].clear();
for (int i = ; i < n; ++ i)
{
int x = inp();
int y = inp();
v[x].push_back(y);
v[y].push_back(x);
}
dfs1(, );
dfs2(, );
for (int x = ; x <= n; ++ x)
{
int ret = ;
int fact = ;
for (int i = ; i <= k; ++ i)
{
(ret += S[k][i]*fact%MOD*g[x][i]) %= MOD;
(fact *= i+) %= MOD;
}
printf("%d\n", (ret+MOD)%MOD);
}
} return ;
}

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