AC自动机

放三个板子

板子1:

#include<bits/stdc++.h>
using namespace std;
const int N=2e6;
int tire[N][26],fail[N],n,tot=1,end[N],ans;
char str[N];
inline void insert(char *a)
{
    int n=strlen(a+1),p=1;
    for(int i=1;i<=n;++i)
    {
        int ch=a[i]-'a';
        if(!tire[p][ch]) tire[p][ch]=++tot;
        p=tire[p][ch];
    }
    end[p]++;
}
inline void get_fail()
{
    queue<int>q;q.push(1);
    fail[1]=0;
    for(int i=0;i<26;++i) tire[0][i]=1;
    while(!q.empty())
    {
        int x=q.front();q.pop();
        for(int i=0;i<26;++i)
        {
            if(!tire[x][i]) {tire[x][i]=tire[fail[x]][i];continue;}
            fail[tire[x][i]]=tire[fail[x]][i];q.push(tire[x][i]); 
        }
    }
}
inline int AC(char *a)
{
    int n=strlen(a+1),p=1;
    for(int i=1;i<=n;++i)
    {
        int ch=a[i]-'a';
        int k=tire[p][ch];
        while(k>1&&end[k]!=-1)
        {
            ans+=end[k];end[k]=-1;
            k=fail[k];
        }
        p=tire[p][ch];
    }
    return ans;
}
int main()
{
    freopen("1.in","r",stdin);
    scanf("%d",&n);
    for(int i=1;i<=n;++i)
    {
        scanf("%s",str+1);
        insert(str);
    }
    get_fail();
    scanf("%s",str+1);
    printf("%d",AC(str));
    return 0;
} 
View Code

板子2:

#include<bits/stdc++.h>
using namespace std;
const int N=1e5,qwq=1e6+10;
int n,tire[N][26],tot=1,end[N],cnt[N],fail[N];
char str[200][100];
char ch[qwq];
inline void insert(char *a,int s)
{
    int n=strlen(a+1),p=1;
    for(int i=1;i<=n;++i)
    {
        int ch=a[i]-'a';
        if(!tire[p][ch]) tire[p][ch]=++tot;
        p=tire[p][ch];
    }
    end[p]=s;
}
inline void get_fail()
{
    queue<int>q;q.push(1);
    fail[1]=0;
    for(int i=0;i<26;++i) tire[0][i]=1;
    while(!q.empty())
    {
        int x=q.front();q.pop();
        for(int i=0;i<26;++i)
        {
            if(!tire[x][i]) {tire[x][i]=tire[fail[x]][i];continue;}
            fail[tire[x][i]]=tire[fail[x]][i];q.push(tire[x][i]);
        }
    }
}
inline void AC(char *a)
{
    int m=strlen(a+1),p=1;
    for(int i=1;i<=m;++i)
    {
        int ch=a[i]-'a';
        int k=tire[p][ch];
        while(k>1)
        {
            cnt[end[k]]++;
            k=fail[k];
        }
        p=tire[p][ch];
        //cout<<i<<' '<<cnt[1]<<' '<<cnt[2]<<endl;
    }
    int mx=0;
    for(int i=1;i<=n;++i) mx=max(mx,cnt[i]);
    printf("%d\n",mx);
    for(int i=1;i<=n;++i)
        if(cnt[i]==mx) printf("%s\n",str[i]+1);
}
int main()
{
    freopen("1.in","r",stdin);
    while(scanf("%d",&n))
    {
        if(n==0) break;
        memset(tire,0,sizeof(tire));
        memset(fail,0,sizeof(fail));
        memset(end,0,sizeof(end));
        memset(cnt,0,sizeof(cnt));
        tot=1;
        for(int i=1;i<=n;++i)
        {
            scanf("%s",str[i]+1);
            insert(str[i],i);
        }
        get_fail();
        scanf("%s",ch+1);
        AC(ch);
    }
    return 0;
}
View Code

板子3(拓扑优化):

#include<bits/stdc++.h>
using namespace std;
const int N=2e6;
int tire[N][26],fail[N],n,tot=1,end[N],ans;
char str[N];
inline void insert(char *a)
{
    int n=strlen(a+1),p=1;
    for(int i=1;i<=n;++i)
    {
        int ch=a[i]-'a';
        if(!tire[p][ch]) tire[p][ch]=++tot;
        p=tire[p][ch];
    }
    end[p]++;
}
inline void get_fail()
{
    queue<int>q;q.push(1);
    fail[1]=0;
    for(int i=0;i<26;++i) tire[0][i]=1;
    while(!q.empty())
    {
        int x=q.front();q.pop();
        for(int i=0;i<26;++i)
        {
            if(!tire[x][i]) {tire[x][i]=tire[fail[x]][i];continue;}
            fail[tire[x][i]]=tire[fail[x]][i];q.push(tire[x][i]); 
        }
    }
}
inline int AC(char *a)
{
    int n=strlen(a+1),p=1;
    for(int i=1;i<=n;++i)
    {
        int ch=a[i]-'a';
        int k=tire[p][ch];
        while(k>1&&end[k]!=-1)
        {
            ans+=end[k];end[k]=-1;
            k=fail[k];
        }
        p=tire[p][ch];
    }
    return ans;
}
int main()
{
    freopen("1.in","r",stdin);
    scanf("%d",&n);
    for(int i=1;i<=n;++i)
    {
        scanf("%s",str+1);
        insert(str);
    }
    get_fail();
    scanf("%s",str+1);
    printf("%d",AC(str));
    return 0;
}
View Code

 

posted @ 2020-04-20 19:29  逆天峰  阅读(132)  评论(0编辑  收藏  举报
作者:逆天峰
出处:https://www.cnblogs.com/gcfer//