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

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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<set>#include<cmath>#include<cstdio>#include<cstring>#include<algorithm>#define LD double#define LL long long#define ULL unsigned long long#define min(a,b) ((a<b)?a:b)#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;}

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