【题解】Luogu P4052[JSOI2007]文本生成器 AC自动机

AC自动机上DP

f[i][j]表示节点j,串长为i

当一个串的尾节点被标红或其fail指针指向的被标红，都是可读的

用总的减去不可读的即为答案

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#define MOD (10007)
#define  N  (10005)
using namespace std;int Son[N][26],End[N],Fail[N];
int n,m,sz,f[105][N],ans;
char s[N];
queue<int>q;void Insert(char s[])
{int now=0,len=strlen(s);for (int i=0; i<len; ++i){int x=s[i]-'A';if (!Son[now][x]) Son[now][x]=++sz;now=Son[now][x];}End[now]|=1;
}void Build_Fail()
{for (int i=0; i<26; ++i)if (Son[0][i])q.push(Son[0][i]);while (!q.empty()){int now=q.front();q.pop();for (int i=0; i<26; ++i){if (!Son[now][i]){Son[now][i]=Son[Fail[now]][i];continue;}End[Son[now][i]]|=End[Son[Fail[now]][i]];Fail[Son[now][i]]=Son[Fail[now]][i];q.push(Son[now][i]);}}
}int main()
{scanf("%d%d",&n,&m);for (int i=1; i<=n; ++i)scanf("%s",s),Insert(s);Build_Fail();f[0][0]=1;for (int i=1; i<=m; ++i)for (int j=0; j<=sz; ++j)for (int k=0; k<26; ++k)if (!End[Son[j][k]])(f[i][Son[j][k]]+=f[i-1][j])%=MOD;for (int i=0; i<=sz; ++i)(ans+=f[m][i])%=MOD;int sum=1;for (int i=1; i<=m; ++i)sum=sum*26%MOD;printf("%d\n",(sum-ans+MOD)%MOD);
}