洛谷 - P4052 文本生成器

传送门：洛谷 - P4052

思路分析

求至少包含一个单词的串不好求，可以求一个都不包含的，然后用总数减去
如果一个点的fail包含，那么这个点也包含，标记一下
$dp[i][j]$表示匹配前i个字母，当前在ac自动机的第j个节点，一个都不包含的方案数

AC代码

#include <bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (int)(x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

namespace FastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
    bool IOerror=0;
    inline char nc() {
        static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
        if(p1==pend) {
            p1=buf;
            pend=buf+fread(buf,1,BUF_SIZE,stdin);
            if(pend==p1) {
                IOerror=1;
                return -1;
            }
        }
        return *p1++;
    }
    inline bool blank(char ch) {
        return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';
    }
    template<class T> inline bool read(T &x) {
        bool sign=0;
        char ch=nc();
        x=0;
        for(; blank(ch); ch=nc());
        if(IOerror)return false;
        if(ch=='-')sign=1,ch=nc();
        for(; ch>='0'&&ch<='9'; ch=nc())x=x*10+ch-'0';
        if(sign)x=-x;
        return true;
    }
    template<class t> inline void print(t x) {
        if (x < 0) putchar('-'), print(-x);
        else {
            if (x > 9) print(x / 10);
            putchar('0' + x % 10);
        }
    }
    template<class T,class... U>bool read(T& h,U&... t) {
        return read(h)&&read(t...);
    }
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace std;
using namespace FastIO;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

typedef long long ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;

const int INF = 0x3f3f3f3f;
const int N = 1e6+10;
const int mod = 1e4+7;


struct AC_auto {
    struct T {
        int son[26],fail,word,last;
        void clear() {
            mem(son,0);
            word = fail = last = 0;
        }
    } tree[N];

    int cnt = 1;
    int vis[N] ;
    int dp[110][100010];
    
    int newnode() {
        ++cnt;
        tree[cnt].clear();
        return cnt;
    }

    void init(){
        cnt = 1;
        tree[1].clear();
        tree[1].word = 1;
    }

    void ins(string s) {
        int p = 1;
        for(auto i:s) {
            int c = i-'A';
            if(!tree[p].son[c]) tree[p].son[c] = newnode();
            p = tree[p].son[c];
        }
        tree[p].word = 1;
        vis[p] = 1;
    }

    void getfail() {
        for(int i=0; i<26; i++) tree[0].son[i] = 1;
        queue<int>q;
        q.push(1);
        while(!q.empty()) {
            int u = q.front();
            q.pop();
            int ufail = tree[u].fail;
            tree[u].last = tree[ufail].word>0?ufail:tree[ufail].last;
            for(int i=0; i<26; i++) {
                int v = tree[u].son[i];
                if(v) {
                    tree[v].fail = tree[ufail].son[i];
                    vis[v]|=vis[tree[v].fail];
                    q.push(v);
                } else tree[u].son[i] = tree[ufail].son[i];
            }
        }
    }

    int solve(int m){
        dp[0][1] = 1;
        for(int i=1;i<=m;i++){
            for(int j=1;j<=cnt;j++){
                for(int k=0;k<26;k++){
                    int v = tree[j].son[k];
                    if(!vis[v]) dp[i][v] = (dp[i][v]+dp[i-1][j])%mod;
                }
            }
        }
        int res = 1;
        for(int i=1;i<=m;i++) res = (res*26)%mod;
        for(int i=1;i<=cnt;i++) res = (res - dp[m][i] + mod) %mod;
        return res;
    }
    
}AC; 


signed main() {

#ifdef xiaofan
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
#endif

    int n,m;
    cin>>n>>m;
    AC.init();
    for(int i=1;i<=n;i++){
        string s;
        cin>>s;
        AC.ins(s);
    } 
    AC.getfail();
    cout<<AC.solve(m)<<endl;





    return 0;
}