# 4730 Matches

Title Description

Tree size $ n $, $ I $ first side of a character set $ S_i $. $ M $ a given pattern string $ t_1, t_2, \ dots, t_m $.

$ Q $ times asking $ (u, v) $, provided $ u \ to v $ passing side is $ e_1, e_2, \ dots, e_k $, find the number of the string $ s $ of solutions to meet:
- $ | S | = K $
- $ \ FORALL I \ in [. 1, K], S_I \ in e_i of S_ {} $
- $ \ J exist, t_j \ text {IS} of the substring S $ A

answer

Consider the total number of illegal program subtracts the number of programs to build $ \ text {AC} $ automata, i.e. $ \ text {dp} $: $ f [i] [j] $ $ I $ represents the front characters, currently the number of programs on the $ j $ node numbers on an automatic machine, consider each step can not come to the end node. And then found can be written in matrix form, written on the test block approach also can not seem to make life difficult for optimization, so we can pre-$ nlogn $ matrix, then every time you jump up, you can use a matrix vector multiply, so efficiency It is $ O (40 ^ 3nlogn + 40 ^ 2Qlogn) $, go through, in that the preprocessing section bottleneck.

Consider alternative hop $ \ text {lca} $ manner: each transition $ \ le lowbit (dp [x]) $ step, this jump number will not exceed $ O (logn) $ and for each point only require pretreatment $ log (lowbit (dp [x])) + 1 $ matrices can.

So we for $ \ text {deep} $ each bit, if the number on the first $ I $ site $ 1 $ to less than $ 0 $, then, we have all the $ \ text {deep} $ plus $ 2 ^ i $ to, the maximum number of such pretreatment is $ \ frac {n} {2} \ times 1+ \ frac {n} {4} \ times 2+ \ frac {n} {8} \ times 3+ ... <2n $, so that efficiency is $ O (40 ^ 3n + 40 ^ 2Qlogn) $.

Code

#include <bits/stdc++.h>
using namespace std;
const int N=5005,M=42,P=998244353;
int n,m,q,hd[N],V[N],nx[N],t=1,e[M],tr[M][M],fi[M],fa[N][13],dp[N],Y,Z,f[2][M],su[2],c[N],b[N];
char g[N][M],h[M];queue<int>qu;bool F[N];
struct O{
    int p[M][M];
}d[M],a[N],G,up[2501][13],dn[2501][13],S[100];
void add(int u,int v){
    nx[++t]=hd[u];V[hd[u]=t]=v;
}
int X(int x){return x>=P?x-P:x;}
void ins(){
    int p=0,l=strlen(h);
    for (int k,i=0;i<l;i++){
        k=h[i]-'a';
        if (!tr[p][k])
            tr[p][k]=++t;
        p=tr[p][k];
    }
    e[p]=1;
}
void build(){
    for (int i=0;i<26;i++)
        if (tr[0][i]) qu.push(tr[0][i]);
    while(!qu.empty()){
        int k=qu.front();qu.pop();
        for (int i=0;i<26;i++)
            if (tr[k][i])
                fi[tr[k][i]]=tr[fi[k]][i],
                e[tr[k][i]]=e[tr[k][i]]|e[fi[tr[k][i]]],
                qu.push(tr[k][i]);
            else tr[k][i]=tr[fi[k]][i];
    }
}
O Add(O A,O B){
    for (int i=0;i<=t;i++)
        for (int j=0;j<=t;j++)
            A.p[i][j]=X(A.p[i][j]+B.p[i][j]);
    return A;
}
O Mul(O A,O B){
    for (int i=0;i<=t;i++)
        for (int j=0;j<=t;j++) G.p[i][j]=0;
    for (int k=0;k<=t;k++)
        for (int j=0;j<=t;j++) if (A.p[k][j])
            for (int i=0;i<=t;i++) if (B.p[i][k])
                G.p[i][j]=X(G.p[i][j]+1ll*A.p[k][j]*B.p[i][k]%P);
    return G;
}
void dfs(int u,int fr){
    dp[u]=dp[fa[u][0]=fr]+1;
    for (int v,j,i=hd[u];i;i=nx[i]){
        if ((v=V[i])==fr) continue;
        j=i>>1;b[v]=strlen(g[j]);
        for (int k=0;k<b[v];k++)
            a[v]=Add(a[v],d[g[j][k]-97]);
        dfs(v,u);
    }
}
void dfs(int u){
    for (int v,i=hd[u],w;i;i=nx[i]){
        if ((v=V[i])==fa[u][0]) continue;
        w=c[dp[v]&-dp[v]];
        up[v][0]=dn[v][0]=a[v];
        for (int j=1;j<=w;j++)
            up[v][j]=Mul(up[v][j-1],up[fa[v][j-1]][j-1]),
            dn[v][j]=Mul(dn[fa[v][j-1]][j-1],dn[v][j-1]);
        dfs(v);
    }
}
int lca(int u,int v,int &w){
    if (dp[u]<dp[v]) swap(u,v);
    while(dp[u]>dp[v]) w=1ll*w*b[u]%P,u=fa[u][0];
    while(u!=v) w=1ll*w*b[u]%P,
        w=1ll*w*b[v]%P,u=fa[u][0],v=fa[v][0];
    return u;
}
void Dp(O A){
    for (int i=0;i<=t;i++){
        f[Y][i]=0;
        for (int j=0;j<=t;j++) if (A.p[i][j])
            f[Y][i]=X(f[Y][i]+1ll*A.p[i][j]*f[Z][j]%P);
    }
    Z^=1;Y^=1;
}
int main(){
    cin>>n>>m>>q;
    for (int u,v,i=1;i<n;i++)
        scanf("%d%d%s",&u,&v,g[i]),
        add(u,v),add(v,u);t=0;
    for (int i=1;i<=m;i++)
        scanf("%s",h),ins();build();
    for (int i=0;i<26;i++)
        for (int k,j=0;j<=t;j++)
            if (!e[k=tr[j][i]]) d[i].p[k][j]++;
    dfs(1,0);fa[1][0]=1;c[1<<12]=12;
    for (int i=0;i<12;i++){
        su[0]=su[1]=0;
        for (int j=1;j<=n;j++)
            su[(dp[j]>>i)&1]++;
        if (su[0]>su[1]){
            for (int j=1;j<=n;j++)
                dp[j]+=(1<<i);
        }
        c[1<<i]=i;
    }
    for (int i=0;i<13;i++)
        for (int j=0;j<=t;j++)
            up[1][i].p[j][j]=dn[1][i].p[j][j]=1;
    for (int i=1;i<13;i++)
        for (int j=1;j<=n;j++)
            fa[j][i]=fa[fa[j][i-1]][i-1];
    dfs(1);
    for (int x,y,w,u,v,z,o;q--;){
        scanf("%d%d",&x,&y);
        Y=f[o=Z=0][0]=w=1;
        z=lca(x,y,w);
        while(x!=z){
            u=dp[x]&-dp[x];
            for (int i=u;i;i>>=1)
                if (dp[x]-i>=dp[z]){
                    S[++o]=up[x][c[i]];
                    x=fa[x][c[i]];break;
                }
        }
        v=o;
        while(y!=z){
            u=dp[y]&-dp[y];
            for (int i=u;i;i>>=1)
                if (dp[y]-i>=dp[z]){
                    S[++o]=dn[y][c[i]];
                    y=fa[y][c[i]];break;
                }
        }
        for (int i=1;i<=v;i++) Dp(S[i]);
        for (int i=o;i>v;i--) Dp(S[i]);
        for (int i=0;i<=t;i++)
            w=X(w+P-f[Z][i]),f[0][i]=f[1][i]=0;
        printf("%d\n",w);
    }
    return 0;
}