HDU 4625. JZPTREE

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：

由斯特林数的性质，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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HDU 4625. JZPTREE

解1：

解2：

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