【jzoj5078】【GDOI2017第三轮模拟day2】【魔法咒语】【ac自动机】【矩阵快速幂】

题目大意

解题思路

对非法串构ac自动机，对于l较小的情况，设f[i][j]表示长度为i,在ac自动机上j点的方案数，直接dp即可。

对于加入串len<=2的情况，对于每个j拆成两个点矩阵快速幂即可。

code

#include
#include
#include
#include
#include
#define LD double
#define LL long long
#define ULL unsigned long long
#define min(a,b) ((a
#define max(a,b) ((a>b)?a:b)
#define fo(i,j,k) for(int i=j;i<=k;i++)
#define fd(i,j,k) for(int i=j;i>=k;i--)
#define fr(i,j) for(int i=begin[j];i;i=next[i])
using namespace std;
int const mn=100+9,mm=4*1e5+9,mo=1e9+7,inf=1e9;
int n,m,l,pon,tag[mn],fail[mn],q[mn],map[mn][30],f[mn][mn],ans[209][209],
    mat[209][209],tmp[209][209];
char s1[mn][mn],s[mn];
void multansmat(){
    fo(i,0,2*pon+1)fo(j,0,2*pon+1)tmp[i][j]=0;
    fo(i,0,2*pon+1)fo(j,0,2*pon+1)if(ans[i][j])fo(k,0,2*pon+1)tmp[i][k]=(tmp[i][k]+1ll*ans[i][j]*mat[j][k])%mo;
    fo(i,0,2*pon+1)fo(j,0,2*pon+1)ans[i][j]=tmp[i][j];
}
void multmatmat(){
    fo(i,0,2*pon+1)fo(j,0,2*pon+1)tmp[i][j]=0;
    fo(i,0,2*pon+1)fo(j,0,2*pon+1)if(mat[i][j])fo(k,0,2*pon+1)tmp[i][k]=(tmp[i][k]+1ll*mat[i][j]*mat[j][k])%mo;
    fo(i,0,2*pon+1)fo(j,0,2*pon+1)mat[i][j]=tmp[i][j];
}
int main(){
    //freopen("sorcery.in","r",stdin);
    //freopen("sorcery.out","w",stdout);
    freopen("d.in","r",stdin);
    freopen("d.out","w",stdout);
    scanf("%d%d%d\n",&n,&m,&l);
    fo(i,1,n)scanf("%s\n",s1[i]+1);
    fo(cas,1,m){
        scanf("%s\n",s+1);
        int now=0,pre,len=strlen(s+1);
        fo(i,1,len)
            now=map[now][s[i]-'a']=(map[now][s[i]-'a'])?map[now][s[i]-'a']:++pon;
        tag[now]=1;
    }
    int ti=0;
    fo(i,0,25)if(map[0][i])q[++ti]=map[0][i];
    fo(i,1,pon){
        fo(j,0,25)if(map[q[i]][j]){
            int next=map[q[i]][j];
            fail[next]=fail[q[i]];
            while(fail[next]&&(!map[fail[next]][j]))
                tag[next]|=tag[fail[next]],
                fail[next]=fail[fail[next]];
            fail[next]=map[fail[next]][j];
            q[++ti]=next;
            tag[next]|=tag[q[i]];
            tag[next]|=tag[fail[next]];
        }
    }
    if(l<=100){
        f[0][0]=1;
        fo(i,0,l)fo(j,0,pon)if(f[i][j])fo(cas,1,n){
            int now=j,ok=!tag[j],len=strlen(s1[cas]+1);
            fo(k,1,len){
                while(now&&(!map[now][s1[cas][k]-'a']))now=fail[now];
                now=map[now][s1[cas][k]-'a'];
                if(tag[now]){ok=0;break;}
            }
            if(ok)f[i+len][now]=(f[i+len][now]+f[i][j])%mo;
        }
        int ans=0;
        fo(j,0,pon)ans=(ans+f[l][j])%mo;
        printf("%d",ans);
    }else{
        ans[1][pon+1]=1;
        fo(cas,1,n){
            int now,ok,len=strlen(s1[cas]+1);
            fo(i,0,pon)if(!tag[i]){
                now=i;ok=1;
                fo(j,1,len){
                    while(now&&(!map[now][s1[cas][j]-'a']))now=fail[now];
                    now=map[now][s1[cas][j]-'a'];
                    if(tag[now]){ok=0;break;}
                }
                if(len==1)mat[i+pon+1][now+pon+1]+=ok;
                else mat[i][now+pon+1]+=ok;
            }
        }
        fo(i,0,pon)mat[i+pon+1][i]++;
        while(l){
            if(l&1)multansmat();
            multmatmat();
            l>>=1;
        }
        int anss=0;
        fo(j,0,pon)anss=(anss+ans[1][j+pon+1])%mo;
        printf("%d",anss);
    }
    return 0;
}