题目大意
https://www.luogu.org/problemnew/show/CF997D
分析
我们发现两棵树互不相关
于是我们可以分别求出两棵树的信息
我们点分,人啊按后设f[i][x]为从根出发走i步到x中间不经过根的方案数,g[i][x]为可以经过根的方案数
代码
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<ctime>
#include<vector>
#include<set>
#include<map>
#include<stack>
using namespace std;
#define add(x,y) x=(x+y)%mod
const int mod = 998244353;
int c[80][80],m;
struct tree {
int n,ans[80],f[80][5000],g[80][5000];
int root,sum,siz[5000],w[5000],a[5000],tot;
bool vis[5000];
vector<int>v[5000];
inline void getrt(int x,int fa){
siz[x]=1;
w[x]=0;
for(int i=0;i<v[x].size();i++)
if(v[x][i]!=fa&&!vis[v[x][i]]){
getrt(v[x][i],x);
siz[x]+=siz[v[x][i]];
w[x]=max(w[x],siz[v[x][i]]);
}
w[x]=max(w[x],sum-siz[x]);
if(!root||w[x]<w[root])root=x;
}
inline void dfs(int x,int fa){
a[++tot]=x;
siz[x]=1;
for(int i=0;i<v[x].size();i++)
if(v[x][i]!=fa&&!vis[v[x][i]]){
dfs(v[x][i],x);
siz[x]+=siz[v[x][i]];
}
}
inline void work(int x){
tot=0;
dfs(x,0);
int i,j,k;
memset(f,0,sizeof(f));
memset(g,0,sizeof(g));
f[0][x]=g[0][x]=1;
for(i=1;i<=m;i++)
for(j=1;j<=tot;j++){
for(k=0;k<v[a[j]].size();k++){
if(vis[v[a[j]][k]])continue;
if(a[j]!=x)add(f[i][a[j]],f[i-1][v[a[j]][k]]);
add(g[i][a[j]],g[i-1][v[a[j]][k]]);
}
}
for(i=1;i<=tot;i++){
if(a[i]==x){
for(j=0;j<=m;j++)add(ans[j],g[j][a[i]]);
}else {
for(j=0;j<=m;j++)
for(k=0;k+j<=m;k++)
add(ans[k+j],(long long)f[j][a[i]]*g[k][a[i]]%mod);
}
}
vis[x]=1;
for(i=0;i<v[x].size();i++)
if(!vis[v[x][i]]){
root=0;
sum=siz[v[x][i]];
getrt(v[x][i],x);
work(root);
}
}
inline void solve(){
sum=n;
root=0;
memset(vis,0,sizeof(vis));
memset(ans,0,sizeof(ans));
getrt(1,0);
work(root);
}
};
tree t1,t2;
int main(){
int i,j,k;
scanf("%d%d%d",&t1.n,&t2.n,&m);
c[0][0]=1;
for(i=1;i<=m;i++)c[i][i]=c[i][0]=1;
for(i=1;i<=m;i++)
for(j=1;j<i;j++)c[i][j]=(c[i-1][j]+c[i-1][j-1])%mod;
for(i=0;i<=t1.n;i++)t1.v[i].clear();
for(i=0;i<=t2.n;i++)t2.v[i].clear();
for(i=1;i<t1.n;i++){
int x,y;
scanf("%d%d",&x,&y);
t1.v[x].push_back(y);
t1.v[y].push_back(x);
}
for(i=1;i<t2.n;i++){
int x,y;
scanf("%d%d",&x,&y);
t2.v[x].push_back(y);
t2.v[y].push_back(x);
}
t1.solve(),t2.solve();
int Ans=0;
for(i=0;i<=m;i++)
Ans=(Ans+(long long)t1.ans[i]*t2.ans[m-i]%mod*c[m][i]%mod)%mod;
cout<<Ans;
return 0;
}