题面

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Sol

先建立AC自动机，trie树上每个字符串结束的位置记录下它的长度\(len\)
设\(f[i]\)表示前\(i\)个字符是否被翻译
在AC自动机上匹配，跳\(fail\)转移，该点为\(x\)，则\(f[i]\)由\(f[i-len[x]]\)转移而来
太菜了不会trie的暴力

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(4e6 + 10);

IL ll Read(){
    RG ll x = 0, z = 1; RG char c = getchar();
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    return x * z;
}

int n, m, tot, ch[26][23333], ed[23333], fail[23333];
bool f[_];
char s[30], t[_];
queue <int> Q;

IL void Insert(){
    RG int len = strlen(s), x = 0;
    for(RG int i = 0; i < len; ++i){
        if(!ch[s[i] - 'a'][x]) ch[s[i] - 'a'][x] = ++tot;
        x = ch[s[i] - 'a'][x];
    }
    ed[x] = len;
}

IL void Get_Fail(){
    for(RG int i = 0; i < 26; ++i) if(ch[i][0]) Q.push(ch[i][0]);
    while(!Q.empty()){
        RG int fa = Q.front(); Q.pop();
        for(RG int i = 0; i < 26; ++i)
            if(!ch[i][fa]) ch[i][fa] = ch[i][fail[fa]];
            else fail[ch[i][fa]] = ch[i][fail[fa]], Q.push(ch[i][fa]);
    }
}

IL int Compare(){
    Fill(f, 0); f[0] = 1; RG int ans = 0, len = strlen(t + 1);
    for(RG int i = 1, x = 0; i <= len; ++i){
        x = ch[t[i] - 'a'][x];
        for(RG int j = x; j; j = fail[j]){
            f[i] |= f[i - ed[j]];
            if(f[i]) break;
        }
    }
    for(RG int i = 1; i <= len; ++i) if(f[i]) ans = i;
    return ans;
}

int main(RG int argc, RG char* argv[]){
    n = Read(); m = Read();
    for(RG int i = 1; i <= n; ++i) scanf(" %s", s), Insert();
    Get_Fail();
    for(RG int i = 1; i <= m; ++i) scanf(" %s", t + 1), printf("%d\n", Compare());
    return 0;
}