AC自动机例题

P3808 [模板]AC自动机(简单版)
[题目描述]
给定n个模式串和1个文本串，求有多少个模式串在文本串里出现过。

#include
using namespace std;
typedef long long LL;
const int INF=1e9+7;
inline LL read(){
    register LL x=0,f=1;register char c=getchar();
    while(c<48||c>57){if(c=='-')f=-1;c=getchar();}
    while(c>=48&&c<=57)x=(x<<3)+(x<<1)+(c&15),c=getchar();
    return f*x;
}

const int MAXN=1e6+5;
const int MAXV=26;

namespace ACzdj{
    struct Trie{
        int v[MAXV];
        int fail,end;
    }AC[MAXN];
    int Ncnt;
    inline void build(string s){
        int cur=0,l=s.length();
        for(int i=0;i q;
        for(int i=0;i<26;i++)
            if(AC[0].v[i]){
                AC[AC[0].v[i]].fail=0;
                q.push(AC[0].v[i]);
            }
        while(!q.empty()){
            int u=q.front();q.pop();
            for(int i=0;i<26;i++){
                if(AC[u].v[i]){
                    AC[AC[u].v[i]].fail=AC[AC[u].fail].v[i];
                    q.push(AC[u].v[i]);
                }
                else
                    AC[u].v[i]=AC[AC[u].fail].v[i];
            }
        }
    }
    inline int query(string s){
        int res=0;
        int l=s.length(),cur=0;
        for(int i=0;i>s;
        build(s);
    }
    AC[0].fail=0;
    Get_fail();
    cin>>s;
    printf("%d\n",query(s));
}

P3796 [模板]AC自动机(加强版)
[题目描述]
有NN个由小写字母组成的模式串以及一个文本串TT。每个模式串可能会在文本串中出现多次。你需要找出哪些模式串在文本串TT中出现的次数最多。

// luogu-judger-enable-o2
#include
using namespace std;
typedef long long LL;
const int INF=1e9+7;
inline LL read(){
    register LL x=0,f=1;register char c=getchar();
    while(c<48||c>57){if(c=='-')f=-1;c=getchar();}
    while(c>=48&&c<=57)x=(x<<3)+(x<<1)+(c&15),c=getchar();
    return f*x;
}

const int MAXN=150000;
const int MAXM=155;
const int MAXV=26;

struct Node{
    int num,id;
    friend bool operator < (Node a,Node b){
        if(a.num==b.num) return a.idb.num;
    }
}ans[MAXM];

namespace ACzdj{
    struct Trie{
        int v[MAXV];
        int fail,end;
    }AC[MAXN];
    int Ncnt;
    inline void clear(int x){
        for (register int i = 0; i < 26; ++i)
            AC[x].v[i] = 0;
        AC[x].fail=0,AC[x].end=0;
    }
    inline void build(const string& s,int id){
        int cur=0,l=s.length();
        for(int i=0;i q;
        for(int i=0;i<26;i++)
            if(AC[0].v[i]){
                AC[AC[0].v[i]].fail=0;
                q.push(AC[0].v[i]);
            }
        while(!q.empty()){
            int u=q.front();q.pop();
            for(int i=0;i<26;i++){
                if(AC[u].v[i]){
                    AC[AC[u].v[i]].fail=AC[AC[u].fail].v[i];
                    q.push(AC[u].v[i]);
                }
                else
                    AC[u].v[i]=AC[AC[u].fail].v[i];
            }
        }
    }
    inline void query(const string& s){
        int l=s.length(),cur=0;
        for(int i=0;i>s[i];
            ans[i].num=0,ans[i].id=i;
            build(s[i],i);
        }
        AC[0].fail=0;
        Get_fail();
        cin>>s[0];
        query(s[0]);
        sort(ans+1,ans+n+1);
        printf("%d\n",ans[1].num);
        cout<

转载于:https://www.cnblogs.com/lizehon/p/10390192.html