[BZOJ4861][Beijing2017]魔法咒語 AC自動機+動態規劃+矩陣快速冪

對忌諱詞語構建AC自動機
Fi,j 表示長度爲i ，匹配到AC自動機第j 位的合法串方案數
當狀態數少的時候用矩陣轉移

#include <bits/stdc++.h>
#define N 5050
const int mod = 1e9+7;
using namespace std;
typedef long long LL;
int len[105],cnt,n,m,l,ans;
char s[55][105];
inline void inc(int &x,int y) {x=(x+y)%mod;}

struct Trie{
    int ch[N][27],leef[N],fail[N];
    char s[N];
    void add() {
        scanf("%s",s+1); int n = strlen(s+1);
        int p = 0;
        for (int i=1;i<=n;i++) {
            int c = s[i] - 'a';
            if (!ch[p][c]) ch[p][c] = ++cnt;
            p = ch[p][c];
        }
        leef[p] = 1;
    }

    void set() {
        queue<int> q;
        for (int i=0;i<26;i++) if (ch[0][i])
            fail[ ch[0][i] ] = 0, q.push(ch[0][i]);

        while (!q.empty()) {
            int u=q.front(); q.pop();
            for (int i=0;i<26;i++) if (ch[u][i]) {
                int v = ch[u][i], t = fail[u];
                while (t && !ch[t][i]) t = fail[t];
                fail[v] = ch[t][i];
                leef[v] = leef[v] || leef[ fail[v] ];
                q.push(v);
            }
        }
    }

    int g(int u,char *s) {
        int len = strlen(s+1);
        for (int i=1;i<=len;i++) {
            int c = s[i]-'a';
            while (u && !ch[u][c]) u = fail[u];
            if (ch[u][c]) u = ch[u][c];
            if (leef[u]) return -1;
        }
        return u;
    }

}T;

namespace planA{
    int F[105][N];
    void solve() {
        F[0][0] = 1;
        for (int _=0;_<l;_++)
            for (int i=0;i<=cnt;i++) if (F[_][i])
                for (int j=1;j<=n;j++) if (_+len[j]<=l) {
                    int ni = T.g(i,s[j]);
                    if (ni == -1) continue;
                    inc(F[_+len[j]][ni], F[_][i]);
                }
        for (int _=0;_<=cnt;_++) inc(ans, F[l][_]);
        cout << ans << endl;
    }
}

namespace planB{

    struct Matrix{int a,d[205][205];}A,B,id;

    Matrix operator*(Matrix p1, Matrix p2) {
        Matrix ret = id;ret.a = p1.a;
        for (int i=0;i<=p1.a;i++)
            for (int j=0;j<=p1.a;j++)
                for (int k=0;k<=p1.a;k++)
                    inc(ret.d[i][j], 1LL*p1.d[i][k] * p2.d[k][j] % mod);
        return ret;
    }

    Matrix qp(Matrix A, int b) {
        Matrix ret = id;
        ret.a = A.a;
        for (int i=0;i<=A.a;i++) ret.d[i][i] = 1;

        while (b) {
            if (b&1) ret = ret * A;
            b >>= 1, A = A * A;
        }
        return ret;
    }

    void solve() {
        int tp = cnt+1;
        A.a = B.a = 2*tp-1;
        A.d[0][0+tp] = 1;

        for (int i=0;i<=cnt;i++) B.d[i+tp][i] = 1;
        for (int i=0;i<=cnt;i++) {
            for (int j=1;j<=n;j++) {
                int ni = T.g(i,s[j]);
                if (ni == -1) continue;
                if (len[j] == 1)
                    B.d[i+tp][ni+tp]++;
                else
                    B.d[i][ni+tp]++;
            }
        }
        A = A * B;
        B = qp(B, l);
        A = A * B;
        for (int i=0;i<=cnt;i++) inc(ans, A.d[0][i]);
        cout << ans << endl;
    }
}

int main() {
    freopen("sorcery.in","r",stdin);
    freopen("sorcery.out","w",stdout);
    scanf("%d%d%d",&n,&m,&l);
    for (int i=1;i<=n;i++) scanf("%s",s[i]+1), len[i] = strlen(s[i]+1);
    for (int i=1;i<=m;i++) T.add();
    T.set();
    if (l <= 100) planA::solve(); else planB::solve();
    return 0;
}