传送门

先将所有模板串扔进广义SAM。发现作文的\(L0\)具有单调性，即\(L0\)更小不会影响答案，所以二分答案。

假设当前二分的值为\(mid\)，将当前的作文放到广义SAM上匹配。

设对于第\(1-i\)个字符来说，最少的失配字符数为\(dp_i\)，那么\(dp_i = dp_{i-1} + 1\)，且如果当前匹配长度\(len \geq mid\)，还有转移\(dp_i = \min\limits_{j=i-len}^{i-mid} dp_j\)。发现在\(i\)增大的过程中\(i-len\)单调不减，因为\(i\)增大\(1\)，\(len\)也最多增大\(1\)。这类似于滑动窗口问题，可以单调队列优化转移，做到复杂度\(O(LlogL)\)。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<ctime>
#include<cctype>
#include<algorithm>
#include<cstring>
#include<iomanip>
#include<queue>
//This code is written by Itst
using namespace std;const int MAXN = 2.2e6 + 7;
namespace SAM{int Lst[MAXN] , Sst[MAXN] , fa[MAXN] , trans[MAXN][2];int cnt = 1 , L;int insert(int p , int len , int x){int t = ++cnt;Lst[t] = len;while(p && !trans[p][x]){trans[p][x] = t;p = fa[p];}if(!p){Sst[t] = fa[t] = 1;return t;}int q = trans[p][x];Sst[t] = Lst[p] + 2;if(Lst[q] == Lst[p] + 1){fa[t] = q;return t;}int k = ++cnt;memcpy(trans[k] , trans[q] , sizeof(trans[k]));Lst[k] = Lst[p] + 1;Sst[k] = Sst[q];Sst[q] = Lst[p] + 2;fa[k] = fa[q];fa[q] = fa[t] = k;while(trans[p][x] == q){trans[p][x] = k;p = fa[p];}return t;}
};
char s[MAXN >> 1];
int ch[MAXN >> 1][2] , dep[MAXN >> 1] , ind[MAXN >> 1];
int N , M , L , cnt = 1;void insert(){scanf("%s" , s + 1);L = strlen(s + 1);int cur = 1;for(int i = 1 ; i <= L ; ++i){if(!ch[cur][s[i] - '0']){ch[cur][s[i] - '0'] = ++cnt;dep[cnt] = dep[cur] + 1;}cur = ch[cur][s[i] - '0'];}
}void build(){queue < int > q;q.push(ind[1] = 1);while(!q.empty()){int x = q.front();q.pop();for(int i = 0 ; i < 2 ; ++i)if(ch[x][i]){ind[ch[x][i]] = SAM::insert(ind[x] , dep[ch[x][i]] , i);q.push(ch[x][i]);}}
}deque < int > q;
int dp[MAXN >> 1];
bool check(int mid){q.clear();int u = 1 , len = 0;for(int i = 1 ; i <= L ; ++i){dp[i] = dp[i - 1] + 1;if(SAM::trans[u][s[i] - '0']){u = SAM::trans[u][s[i] - '0'];++len;}else{while(u && !SAM::trans[u][s[i] - '0'])u = SAM::fa[u];!u ? (u = 1 , len = 0) : (len = SAM::Lst[u] + 1 , u = SAM::trans[u][s[i] - '0']);}while(!q.empty() && q.front() < i - len)q.pop_front();if(len >= mid){while(!q.empty() && dp[q.back()] >= dp[i - mid])q.pop_back();q.push_back(i - mid);}if(!q.empty())dp[i] = min(dp[i] , dp[q.front()]);}return dp[L] <= 0.1 * L;
}int main(){#ifndef ONLINE_JUDGEfreopen("in" , "r" , stdin);//freopen("out" , "w" , stdout);#endifcin >> N >> M;for(int i = 1 ; i <= M ; ++i)insert();build();for(int i = 1 ; i <= N ; ++i){scanf("%s" , s + 1);L = strlen(s + 1);int l = 0 , r = L;while(l < r){int mid = (l + r + 1) >> 1;check(mid) ? l = mid : r = mid - 1;}cout << l << endl;}return 0;
}