传送门并没有
思路
见到那么小的 \(k\) 次方, 又一次想到斯特林数.
\[ ans=\sum_{T} f(T)^k = \sum_{i=0}^k i!S(k,i)\sum_{T} {f(T)\choose i} \]
很套路地, 考虑后面那个式子的组合意义: 对于每一个点集的导出子图, 选出 \(i\) 条边的方案数.
很套路地, 我们想到树形 DP.
设 \(dp_{x,s,0/1}\) 表示 \(x\) 子树内的所有非空点集的导出子图里选出 \(s\) 条边, 点集里有 / 没有 \(x\), 的方案数.
每次加上一棵子树, 就分三种情况考虑: 只有原有的, 只有新的, 合在一起. 其中最后一种要记入答案里.
代码
- #include<bits/stdc++.h>
- clock_t t=clock();
- namespace my_std{
- using namespace std;
- #define pii pair<int,int>
- #define fir first
- #define sec second
- #define MP make_pair
- #define rep(i,x,y) for (int i=(x);i<=(y);i++)
- #define drep(i,x,y) for (int i=(x);i>=(y);i--)
- #define go(x) for (int i=head[x];i;i=edge[i].nxt)
- #define templ template<typename T>
- #define sz 101010
- #define mod 998244353ll
- typedef long long ll;
- typedef double db;
- mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
- templ inline T rnd(T l,T r) {return uniform_int_distribution<T>(l,r)(rng);}
- templ inline bool chkmax(T &x,T y){return x<y?x=y,1:0;}
- templ inline bool chkmin(T &x,T y){return x>y?x=y,1:0;}
- templ inline void read(T& t)
- {
- t=0;char f=0,ch=getchar();double d=0.1;
- while(ch>'9'||ch<'0') f|=(ch=='-'),ch=getchar();
- while(ch<='9'&&ch>='0') t=t*10+ch-48,ch=getchar();
- if(ch=='.'){ch=getchar();while(ch<='9'&&ch>='0') t+=d*(ch^48),d*=0.1,ch=getchar();}
- t=(f?-t:t);
- }
- template<typename T,typename... Args>inline void read(T& t,Args&... args){read(t); read(args...);}
- char sr[1<<21],z[20];int C=-1,Z=0;
- inline void Ot(){fwrite(sr,1,C+1,stdout),C=-1;}
- inline void print(register int x)
- {
- if(C>1<<20)Ot();if(x<0)sr[++C]='-',x=-x;
- while(z[++Z]=x%10+48,x/=10);
- while(sr[++C]=z[Z],--Z);sr[++C]='\n';
- }
- void file()
- {
- #ifndef ONLINE_JUDGE
- freopen("a.in","r",stdin);
- #endif
- }
- inline void chktime()
- {
- #ifndef ONLINE_JUDGE
- cout<<(clock()-t)/1000.0<<'\n';
- #endif
- }
- #ifdef mod
- ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x%mod) if (y&1) ret=ret*x%mod;return ret;}
- ll inv(ll x){return ksm(x,mod-2);}
- #else
- ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x) if (y&1) ret=ret*x;return ret;}
- #endif
- // inline ll mul(ll a,ll b){ll d=(ll)(a*(double)b/mod+0.5);ll ret=a*b-d*mod;if (ret<0) ret+=mod;return ret;}
- }
- using namespace my_std;
- int n,m,K;
- struct hh{int t,nxt;}edge[sz<<1];
- int head[sz],ecnt;
- void make_edge(int f,int t)
- {
- edge[++ecnt]=(hh){t,head[f]};
- head[f]=ecnt;
- edge[++ecnt]=(hh){f,head[t]};
- head[t]=ecnt;
- }
- ll dp[sz][15][2];
- ll f[15][2];
- int size[sz];
- ll ans[15];
- void dfs(int x,int fa)
- {
- dp[x][0][0]=0;dp[x][0][1]=1;++ans[0];
- size[x]=1;
- #define v edge[i].t
- go(x) if (v!=fa)
- {
- dfs(v,x);
- rep(i,0,14) rep(j,0,1) f[i][j]=dp[x][i][j];
- rep(j,0,min(K,size[v])) (f[j][0]+=dp[v][j][0]+dp[v][j][1])%=mod;
- rep(j,0,min(size[x],K))
- {
- rep(k,0,min(size[v],K-j))
- {
- ll S=(dp[v][k][0]+dp[v][k][1])%mod;
- ll s1=dp[x][j][0]*S%mod,s2=dp[x][j][1]*(S+(k?dp[v][k-1][1]:0))%mod;
- (f[j+k][0]+=s1)%=mod,
- (f[j+k][1]+=s2)%=mod;
- (ans[j+k]+=s1+s2)%=mod;
- }
- }
- rep(i,0,K) rep(j,0,1) dp[x][i][j]=f[i][j];
- size[x]+=size[v];
- }
- #undef v
- }
- void solve(){dfs(1,0);}
- ll S[15][15];
- int main()
- {
- file();
- int x,y;
- read(n,m,K);
- rep(i,1,n-1) read(x,y),make_edge(x,y);
- solve();
- S[0][0]=1;
- rep(i,1,K)
- rep(j,1,i)
- S[i][j]=(S[i-1][j-1]+S[i-1][j]*j%mod)%mod;
- ll fac=1,Ans=0;
- rep(i,1,K) fac=fac*i%mod,(Ans+=fac*S[K][i]%mod*ans[i]%mod)%=mod;
- cout<<Ans;
- return 0;
- }
来源: http://www.bubuko.com/infodetail-2969830.html