题意

给你n个串，求长度为m的串的方案数，使得这个串至少包含一个这n个任意一个串

思路

转化为求一个也不包含的方案数
\(dp[i][j]\)为第i个字符，匹配到ac自动机上的j号节点的方案数，显然不能匹配到有结束点的地方
fail节点的性质：指向当前串的最长后缀所在的节点
所以当前节点的fail节点为结束点也不行
所有原本没有的空节点在build的时候会指向根节点
所以空节点都在dp[i][0]里

代码

空间开小了一直wa

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1using namespace std;typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;const db eps = 1e-6;
const int mod = 10007;
const int maxn = 5e5+100;
const int maxm = 2e6+100;
const int inf = 0x3f3f3f3f;
const db pi = acos(-1.0);int dp[111][7777];
int n,m;
struct AC{//局部变量没有默认0!int trie[maxn][26];int num[maxn];//单词出现次数int fail[maxn];int vis[maxn];//ask函数用到int tot;//多测可写个initvoid init(){tot=0;mem(vis,-1);mem(trie,0);}void add(char *s){int root = 0, len = strlen(s+1);for(int i = 1; i <= len; i++){int x = s[i]-'A';if(!trie[root][x])trie[root][x]=++tot;root=trie[root][x];}num[root]++;vis[root]=0;}void build(){queue<int>q;for(int i = 0; i < 26; i++){if(trie[0][i]){fail[trie[0][i]]=0;q.push(trie[0][i]);}}while(!q.empty()){int now = q.front();q.pop();for(int i = 0; i < 26; i++){if(trie[now][i]){fail[trie[now][i]]=trie[fail[now]][i];q.push(trie[now][i]);}else trie[now][i]=trie[fail[now]][i];}vis[now]&=vis[fail[now]];//if(vis[fail[now]]==0)vis[now]=0;}}void solve(){dp[0][0]=1;for(int i = 0; i < m; i++){for(int j = 0; j <= tot; j++){if(vis[j]==0)continue;for(int k = 0; k < 26; k++){int to = trie[j][k];if(vis[to]==0)continue;(dp[i+1][to]+=dp[i][j])%=mod;}}}}}ac;char a[maxn];
int main(){ac.init();scanf("%d %d", &n, &m);for(int i = 1; i <= n; i++){scanf("%s",a+1);ac.add(a);}ac.build();ac.solve();int ans = 1;for(int i = 1; i <= m; i++){ans=ans*26%mod;}/*for(int j = 0; j <= ac.tot; j++){printf("%d %d\n",j,ac.vis[j]);}for(int i = 0; i <= ac.tot; i++){printf("%d:: \n",i);for(int j = 0; j < 26; j++){printf("%d %d\n",j,ac.trie[i][j]);}}*//*for(int i = 1; i <= m; i++){for(int j = 0; j <= ac.tot; j++){printf("%d %d == %d\n",i,j,dp[i][j]);}}*/for(int i = 0; i <= ac.tot; i++){ans=(ans+mod-dp[m][i])%mod;}printf("%d",ans);return 0;
}
/*
10 18
DCSDG
DSSF
SADAV
DSATWYYH
FDHFIS
DFGDFGGD
SASSSS
HHEBB
SFTWRTWW
ZSDFZDS*/