HDU 4625. JZPTREE
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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
$$ E_u = sum_{v=1}^n (d(u, v))^k = sum_{i=0}^k egin{Bmatrix} k \\ i end{Bmatrix} sum_{v=1}^n (d(u, v))^{underline{i}}. $$
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$ ext{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
$$
egin{aligned}
f[u][k]
& = sum_{v in ext{son}(u)} sum_{w in T_v} (d(u, w))^{underline{k}} \\
& = sum_{v in ext{son}(u)} sum_{w in T_v} (d(v, w)+1)^{underline{k}} \\
& = sum_{v in ext{son}(u)} sum_{w in T_v} Big( (d(v, w))^{underline{k}}+k (d(v, w))^{underline{k-1}} Big) \\
& = sum_{v in ext{son}(u)} Big( f[v][k]+k f[v][k-1] Big)
end{aligned}
$$
两遍DFS即可求出所有$E_u$，从而可在$O(nk)$的复杂度内解决。
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 typedef long long ll; 6 typedef unsigned long long ull; 7 typedef double ld; 8 typedef pair<int,int> pii; 9 typedef pair<ll,ll> pll; 10 typedef pair<ld,ld> pdd; 11 12 #define X first 13 #define Y second 14 15 //#include <boost/unordered_map.hpp> 16 //using namespace boost; 17 18 /* 19 #include <ext/pb_ds/tree_policy.hpp> 20 #include <ext/pb_ds/assoc_container.hpp> 21 using namespace __gnu_pbds; 22 typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> rbtree; 23 rbtree T; 24 */ 25 26 namespace io{ 27 const int L = (1 << 20) + 1; 28 29 char buf[L], *S , *T, c; 30 31 char getchar() { 32 if(__builtin_expect(S == T, 0)) { 33 T = (S = buf) + fread(buf, 1, L, stdin); 34 return (S == T ? EOF : *S++); 35 } 36 return *S++; 37 } 38 39 int inp() { 40 int x = 0, f = 1; char ch; 41 for(ch = getchar(); !isdigit(ch); ch = getchar()) 42 if(ch == ‘-‘) f = -1; 43 for(; isdigit(ch); x = x * 10 + ch - ‘0‘, ch = getchar()); 44 return x * f; 45 } 46 47 unsigned inpu() 48 { 49 unsigned x = 0; char ch; 50 for(ch = getchar(); !isdigit(ch); ch = getchar()); 51 for(; isdigit(ch); x = x * 10 + ch - ‘0‘, ch = getchar()); 52 return x; 53 } 54 55 ll inp_ll() { 56 ll x = 0; int f = 1; char ch; 57 for(ch = getchar(); !isdigit(ch); ch = getchar()) 58 if(ch == ‘-‘) f = -1; 59 for(; isdigit(ch); x = x * 10 + ch - ‘0‘, ch = getchar()); 60 return x * f; 61 } 62 63 char B[25], *outs=B+20, *outr=B+20; 64 template<class T> 65 inline void print(register T a,register char x=0){ 66 if(x) *--outs = x, x = 0; 67 68 if(!a)*--outs = ‘0‘; 69 else 70 while(a) 71 *--outs = (a % 10) + 48, a /= 10; 72 73 if(x) 74 *--outs = x; 75 76 fwrite(outs, outr - outs , 1, stdout); 77 outs = outr; 78 } 79 }; 80 81 using io :: print; 82 using io :: inp; 83 using io :: inpu; 84 using io :: inp_ll; 85 86 using i32 = int; 87 using i64 = long long; 88 using u8 = unsigned char; 89 using u32 = unsigned; 90 using u64 = unsigned long long; 91 using f64 = double; 92 using f80 = long double; 93 94 ll power(ll a, ll b, ll p) 95 { 96 if (!b) return 1; 97 ll t = power(a, b/2, p); 98 t = t*t%p; 99 if (b&1) t = t*a%p; 100 return t; 101 } 102 103 ll exgcd(ll a, ll b, ll &x, ll &y) 104 { 105 if (b == 0) 106 { 107 x = 1; 108 y = 0; 109 return a; 110 } 111 ll px, py; 112 ll d = exgcd(b, a%b, px, py); 113 x = py; 114 y = px-a/b*py; 115 return d; 116 } 117 118 template<class T> 119 inline void freshmin(T &a, const T &b) 120 { 121 if (a > b) a = b; 122 } 123 124 template<class T> 125 inline void freshmax(T &a, const T &b) 126 { 127 if (a < b) a = b; 128 } 129 130 const int MAXN = 50010; 131 const int MAXK = 510; 132 const int MOD = 10007; 133 const f80 MI = f80(1)/MOD; 134 const int INF = 100001; 135 136 int n, k; 137 int S[MAXK][MAXK]; 138 139 vector<int> v[MAXN]; 140 int f[MAXN][MAXK], g[MAXN][MAXK]; 141 142 void dfs1(int x, int p) 143 { 144 f[x][0] = 1; 145 for (int i = 1; i <= k; ++ i) 146 f[x][i] = 0; 147 for (auto y : v[x]) 148 { 149 if (y == p) continue; 150 dfs1(y, x); 151 (f[x][0] += f[y][0]) %= MOD; 152 for (int i = 1; i <= k; ++ i) 153 (f[x][i] += f[y][i]+i*f[y][i-1]) %= MOD; 154 } 155 } 156 157 void dfs2(int x, int p) 158 { 159 if (!p) 160 { 161 for (int i = 0; i <= k; ++ i) 162 g[x][i] = f[x][i]; 163 } 164 for (auto y : v[x]) 165 { 166 if (y == p) continue; 167 g[y][0] = g[x][0]; 168 for (int i = 1; i <= k; ++ i) 169 { 170 int g1 = (g[x][i]-(f[y][i]+i*f[y][i-1]))%MOD; 171 int g2 = (g[x][i-1]-(f[y][i-1]+(i-1)*(i-2 >= 0 ? f[y][i-2] : 0)))%MOD; 172 g[y][i] = (f[y][i]+g1+i*g2)%MOD; 173 } 174 dfs2(y, x); 175 } 176 } 177 178 int main() 179 { 180 181 S[0][0] = 1; 182 for (int i = 1; i <= 500; ++ i) 183 for (int j = 1; j <= i; ++ j) 184 S[i][j] = (S[i-1][j-1]+S[i-1][j]*j)%MOD; 185 186 for (int T = inp(); T --; ) 187 { 188 n = inp(); 189 k = inp(); 190 for (int i = 1; i <= n; ++ i) 191 v[i].clear(); 192 for (int i = 1; i < n; ++ i) 193 { 194 int x = inp(); 195 int y = inp(); 196 v[x].push_back(y); 197 v[y].push_back(x); 198 } 199 dfs1(1, 0); 200 dfs2(1, 0); 201 for (int x = 1; x <= n; ++ x) 202 { 203 int ret = 0; 204 for (int i = 0; i <= k; ++ i) 205 (ret += S[k][i]*g[x][i]) %= MOD; 206 printf("%d ", (ret+MOD)%MOD); 207 } 208 } 209 210 return 0; 211 }
解2：
我们用另一个斯特林数的性质：
$$ E_u = sum_{v=1}^n (d(u, v))^k = sum_{i=0}^k i! egin{Bmatrix} k \\ i end{Bmatrix} sum_{v=1}^n inom{d(u, v)}{i}. $$
为此，我们定义
$$f[u][k] = sum_{v in T_u} inom{d(u, v)}{k},$$
其中$T_u$表示以$u$为根节点的子树。令$ ext{son}(u)$表示节点$u$的所有儿子节点的集合，则
$$
egin{aligned}
f[u][k]
& = sum_{v in ext{son}(u)} sum_{w in T_v} inom{d(u, w)}{k} \\
& = sum_{v in ext{son}(u)} sum_{w in T_v} inom{d(v, w)+1}{k} \\
& = sum_{v in ext{son}(u)} sum_{w in T_v} left( inom{d(v, w)}{k} + inom{d(v, w)}{k-1}
ight) \\
& = sum_{v in ext{son}(u)} Big( f[v][k]+f[v][k-1] Big)
end{aligned}
$$
两遍DFS即可求出所有$E_u$，从而可在$O(nk)$的复杂度内解决。
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 typedef long long ll; 6 typedef unsigned long long ull; 7 typedef double ld; 8 typedef pair<int,int> pii; 9 typedef pair<ll,ll> pll; 10 typedef pair<ld,ld> pdd; 11 12 #define X first 13 #define Y second 14 15 //#include <boost/unordered_map.hpp> 16 //using namespace boost; 17 18 /* 19 #include <ext/pb_ds/tree_policy.hpp> 20 #include <ext/pb_ds/assoc_container.hpp> 21 using namespace __gnu_pbds; 22 typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> rbtree; 23 rbtree T; 24 */ 25 26 namespace io{ 27 const int L = (1 << 20) + 1; 28 29 char buf[L], *S , *T, c; 30 31 char getchar() { 32 if(__builtin_expect(S == T, 0)) { 33 T = (S = buf) + fread(buf, 1, L, stdin); 34 return (S == T ? EOF : *S++); 35 } 36 return *S++; 37 } 38 39 int inp() { 40 int x = 0, f = 1; char ch; 41 for(ch = getchar(); !isdigit(ch); ch = getchar()) 42 if(ch == ‘-‘) f = -1; 43 for(; isdigit(ch); x = x * 10 + ch - ‘0‘, ch = getchar()); 44 return x * f; 45 } 46 47 unsigned inpu() 48 { 49 unsigned x = 0; char ch; 50 for(ch = getchar(); !isdigit(ch); ch = getchar()); 51 for(; isdigit(ch); x = x * 10 + ch - ‘0‘, ch = getchar()); 52 return x; 53 } 54 55 ll inp_ll() { 56 ll x = 0; int f = 1; char ch; 57 for(ch = getchar(); !isdigit(ch); ch = getchar()) 58 if(ch == ‘-‘) f = -1; 59 for(; isdigit(ch); x = x * 10 + ch - ‘0‘, ch = getchar()); 60 return x * f; 61 } 62 63 char B[25], *outs=B+20, *outr=B+20; 64 template<class T> 65 inline void print(register T a,register char x=0){ 66 if(x) *--outs = x, x = 0; 67 68 if(!a)*--outs = ‘0‘; 69 else 70 while(a) 71 *--outs = (a % 10) + 48, a /= 10; 72 73 if(x) 74 *--outs = x; 75 76 fwrite(outs, outr - outs , 1, stdout); 77 outs = outr; 78 } 79 }; 80 81 using io :: print; 82 using io :: inp; 83 using io :: inpu; 84 using io :: inp_ll; 85 86 using i32 = int; 87 using i64 = long long; 88 using u8 = unsigned char; 89 using u32 = unsigned; 90 using u64 = unsigned long long; 91 using f64 = double; 92 using f80 = long double; 93 94 ll power(ll a, ll b, ll p) 95 { 96 if (!b) return 1; 97 ll t = power(a, b/2, p); 98 t = t*t%p; 99 if (b&1) t = t*a%p; 100 return t; 101 } 102 103 ll exgcd(ll a, ll b, ll &x, ll &y) 104 { 105 if (b == 0) 106 { 107 x = 1; 108 y = 0; 109 return a; 110 } 111 ll px, py; 112 ll d = exgcd(b, a%b, px, py); 113 x = py; 114 y = px-a/b*py; 115 return d; 116 } 117 118 template<class T> 119 inline void freshmin(T &a, const T &b) 120 { 121 if (a > b) a = b; 122 } 123 124 template<class T> 125 inline void freshmax(T &a, const T &b) 126 { 127 if (a < b) a = b; 128 } 129 130 const int MAXN = 50010; 131 const int MAXK = 510; 132 const int MOD = 10007; 133 const f80 MI = f80(1)/MOD; 134 const int INF = 100001; 135 136 int n, k; 137 int S[MAXK][MAXK]; 138 139 vector<int> v[MAXN]; 140 int f[MAXN][MAXK], g[MAXN][MAXK]; 141 142 void dfs1(int x, int p) 143 { 144 f[x][0] = 1; 145 for (int i = 1; i <= k; ++ i) 146 f[x][i] = 0; 147 for (auto y : v[x]) 148 { 149 if (y == p) continue; 150 dfs1(y, x); 151 (f[x][0] += f[y][0]) %= MOD; 152 for (int i = 1; i <= k; ++ i) 153 (f[x][i] += f[y][i]+f[y][i-1]) %= MOD; 154 } 155 } 156 157 void dfs2(int x, int p) 158 { 159 if (!p) 160 { 161 for (int i = 0; i <= k; ++ i) 162 g[x][i] = f[x][i]; 163 } 164 for (auto y : v[x]) 165 { 166 if (y == p) continue; 167 g[y][0] = g[x][0]; 168 for (int i = 1; i <= k; ++ i) 169 { 170 int g1 = (g[x][i]-(f[y][i]+f[y][i-1]))%MOD; 171 int g2 = (g[x][i-1]-(f[y][i-1]+(i-2 >= 0 ? f[y][i-2] : 0)))%MOD; 172 g[y][i] = (f[y][i]+g1+g2)%MOD; 173 } 174 dfs2(y, x); 175 } 176 } 177 178 int main() 179 { 180 181 S[0][0] = 1; 182 for (int i = 1; i <= 500; ++ i) 183 for (int j = 1; j <= i; ++ j) 184 S[i][j] = (S[i-1][j-1]+S[i-1][j]*j)%MOD; 185 186 for (int T = inp(); T --; ) 187 { 188 n = inp(); 189 k = inp(); 190 for (int i = 1; i <= n; ++ i) 191 v[i].clear(); 192 for (int i = 1; i < n; ++ i) 193 { 194 int x = inp(); 195 int y = inp(); 196 v[x].push_back(y); 197 v[y].push_back(x); 198 } 199 dfs1(1, 0); 200 dfs2(1, 0); 201 for (int x = 1; x <= n; ++ x) 202 { 203 int ret = 0; 204 int fact = 1; 205 for (int i = 0; i <= k; ++ i) 206 { 207 (ret += S[k][i]*fact%MOD*g[x][i]) %= MOD; 208 (fact *= i+1) %= MOD; 209 } 210 printf("%d ", (ret+MOD)%MOD); 211 } 212 } 213 214 return 0; 215 }
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