题面
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;
}