P4827 [国家集训队] Crash 的文明世界

题目描述

Solution

看到这种 $k$ 次幂的式子，就应该往斯特林数的方面想想。

$mn=∑i{ni}(ni)i!m^n=\sum_i \left\{ \begin{aligned} n \\ i \end{aligned} \right\} \left ( \begin{aligned} n \\ i \end{aligned} \right )i!$

因此
$原式$
$=∑i,j{ki}(dist(i,j)i)i!=\sum_{i,j} \left\{ \begin{aligned} k \\ i \end{aligned} \right\} \left ( \begin{aligned} dist(i,j) \\ i \;\;\;\;\;\; \end{aligned} \right )i!$
$=∑i{ki}i!∑j(dist(i,j)i)=\sum_{i} \left\{ \begin{aligned} k \\ i \end{aligned} \right\}i!\sum_j \left ( \begin{aligned} dist(i,j) \\ i \;\;\;\;\;\; \end{aligned} \right )$
$=∑i{ki}i!∑j(dist(i,j)−1i)+∑j(dist(i,j)−1i−1)=\sum_{i} \left\{ \begin{aligned} k \\ i \end{aligned} \right\}i!\sum_j \left ( \begin{aligned} dist(i,j)-1 \\ i\;\;\;\;\;\;\;\; \end{aligned} \right )+\sum_j \left ( \begin{aligned} dist(i,j)-1 \\ i-1 \;\;\;\;\;\; \end{aligned} \right )$

因此设 $F [x] [k]$ 表示在 $x$ 的子树内，距离 $x$ 为 $k$ 的所有节点的贡献。
有：
$F[x][k]=F[son_x][k-1]+F[son_x][k]$
然后换根 $d p$ 即可。
时间复杂度 $O (n k)$ 。

Code

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <cassert>
#include <string.h>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define fi first
#define se secondusing namespace std;template<typename T>inline bool upmin(T &x,T y) { return y<x?x=y,1:0; }
template<typename T>inline bool upmax(T &x,T y) { return x<y?x=y,1:0; }typedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pair<int,int> PR;
typedef vector<int> VI;const lod eps=1e-11;
const lod pi=acos(-1);
const int oo=1<<30;
const ll loo=1ll<<62;
const int mods=10007;
const int MAXN=100005;
const int INF=0x3f3f3f3f;//1061109567
/*--------------------------------------------------------------------*/
inline int read()
{int f=1,x=0; char c=getchar();while (c<'0'||c>'9') { if (c=='-') f=-1; c=getchar(); }while (c>='0'&&c<='9') { x=(x<<3)+(x<<1)+(c^48); c=getchar(); }return x*f;
}
vector<int> e[MAXN];
int f[MAXN][155],S[155][155],fac[155],g[155],n,k;
int upd(int x,int y){ return x+y>=mods?x+y-mods:x+y; }
void tree_dp(int x,int father)
{for (auto v:e[x]){if (v==father) continue;tree_dp(v,x);}f[x][0]=1;for (auto v:e[x]) if (v!=father)for (int i=0;i<=k;i++){f[x][i]=upd(f[x][i],f[v][i]);if (i) f[x][i]=upd(f[x][i],f[v][i-1]);}
}
void dfs(int x,int father)
{for (auto v:e[x]){if (v==father) continue;for (int i=0;i<=k;i++){g[i]=upd(f[x][i],mods-f[v][i]);if (i) g[i]=upd(g[i],mods-f[v][i-1]);}for (int i=0;i<=k;i++){f[v][i]=upd(f[v][i],g[i]);if (i) f[v][i]=upd(f[v][i],g[i-1]);}dfs(v,x);}
}
void Init(int k)
{fac[0]=1;for (int i=1;i<=k;i++) fac[i]=fac[i-1]*i%mods;S[0][0]=1;for (int i=1;i<=k;i++) S[i][i]=S[i][1]=1;for (int i=1;i<=k;i++) for (int j=1;j<i;j++) S[i][j]=upd(S[i-1][j-1],S[i-1][j]*j%mods);
}
int main()
{n=read(),k=read();for (int i=1;i<n;i++){int u=read(),v=read();e[u].PB(v);e[v].PB(u);}tree_dp(1,0);dfs(1,0);Init(k);for (int i=1;i<=n;i++){int ans=0;for (int j=1;j<=k;j++) ans=upd(ans,S[k][j]*fac[j]%mods*f[i][j]%mods); printf("%d\n",ans);}return 0;
}