【模板】AC自动机（加强版）

题目描述

有个由小写字母组成的模式串以及一个文本串。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中出现的次数最多。

输入输出格式

输入格式：

输入含多组数据。

每组数据的第一行为一个正整数，表示共有个模式串，。

接下去行，每行一个长度小于等于的模式串。下一行是一个长度小于等于的文本串。

输入结束标志为。

输出格式：

对于每组数据，第一行输出模式串最多出现的次数，接下去若干行每行输出一个出现次数最多的模式串，按输入顺序排列。

输入输出样例

输入样例#1：

2
aba
bab
ababababac
6
beta
alpha
haha
delta
dede
tata
dedeltalphahahahototatalpha
0

输出样例#1：

4
aba
2
alpha
haha
解法1：。。。。。。。。8813ms

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
struct Tree{int fail,vis[26],end;}AC[100000];//Trie tree
//vis:son_points' position
//end:mark the word_number end with this point
int cnt=0;//Trie's pointer
struct Result{int num,pos;}Ans[100000];//mark every word's appear_times
bool operator <(Result a,Result b){if(a.num!=b.num)return a.num>b.num;else return a.pos<b.pos;}//operate the < operation 
//if it's appear_times higher.just return it,or just return the one have lower pos
string s[100000];
inline void Clean(int x){memset(AC[x].vis,0,sizeof(AC[x].vis));AC[x].fail=0;AC[x].end=0;}
//initialization
inline void Build(string s,int Num){
        int l=s.length(),now=0;//l:mark the length,now:mark the now_position
        for(int i=0;i<l;++i)//make Trie_tree
        {// if the Trie tree has not this point
                if(AC[now].vis[s[i]-'a']==0)AC[now].vis[s[i]-'a']=++cnt,Clean(cnt);
                   //if has not been visited ,just go on
                   //add it to now's son_points and then initialization it 
                now=AC[now].vis[s[i]-'a'];//just run to next level
        }AC[now].end=Num;//mark the end 
}
void Get_fail()//make fail
{queue<int> Q;//set a queue
        for(int i=0;i<26;++i)if(AC[0].vis[i]!=0)//if it has not been visited
                    AC[AC[0].vis[i]].fail=0,Q.push(AC[0].vis[i]);
               //just point it to the root_point and then push it to the queue 
        while(!Q.empty())//BFS to ask for fail
        {int u=Q.front();Q.pop();//use u to mark the queue_head,and then push it out 
              for(int i=0;i<26;++i)
                   if(AC[u].vis[i]!=0)//it it has been visited,just it is real
                      AC[AC[u].vis[i]].fail=AC[AC[u].fail].vis[i],Q.push(AC[u].vis[i]);
                      //make it's son_point's fail against now_fail_son(renew)
                      //and then push then son_point to the queue
                      else AC[u].vis[i]=AC[AC[u].fail].vis[i];
                      //or ,this point has not been visited
                      //just it is not real
                      //just use it's fail(spare tire)
        }
}
int AC_Query(string s){int l=s.length(),now=0,ans=0;
        for(int i=0;i<l;++i){
                now=AC[now].vis[s[i]-'a'];
                for(int t=now;t;t=AC[t].fail)Ans[AC[t].end].num++;
        }return ans;
}
int main(){int n;
     while(1){scanf("%d",&n);if(n==0)break;cnt=0;Clean(0);
         for(int i=1;i<=n;++i)cin>>s[i],Ans[i].num=0,Ans[i].pos=i,Build(s[i],i);
         //read every given_part(key),initialization it
         AC[0].fail=0;Get_fail();
         //initialization root's fail and want fail
         cin>>s[0];AC_Query(s[0]);
         //read the given_long_word and ask for the ans
         sort(&Ans[1],&Ans[n+1]);printf("%d
",Ans[1].num);
         //sort the ans and printf the biggest solution
         cout<<s[Ans[1].pos]<<endl;//give the appear_most_often_word
         for(int i=2;i<=n;++i)if(Ans[i].num==Ans[i-1].num)cout<<s[Ans[i].pos]<<endl;else break;
         //then if have more than one solution, just printf them
         //after sort ,the appear_most_often_word must be in the front
         //and the Ans[1] has printf just start from 2
         //if it is not the best solution,it means there is no best solution any more,just break 
     }return 0;
}//this program's speed is very slow,can only accept just in time 
//maybe we should search for a better solution

虽然实现的很自然，但是这效率。。。。。。光荣地拿到了luogu rank 倒数第一

能有更好的解法吗？

自然是有的。

解法2：1688ms最慢点464ms

无注解，清新模板版

#include<queue>
#include<cstdio>
#include<cstring>
#define maxn 15000
#define maxc 1000005
#define For(a,b,c) for(a=b;a<=(int)(c);++a)
using namespace std;
int n;
char T[maxc],P[151][71];
inline int idx(char c){return c-'a';}
struct Trie{int sz,ch[maxn][26],val[maxn];
    inline void init(){sz=0;memset(ch,0,sizeof(ch));memset(val,0,sizeof(val));}
    inline void insert(char *s,int v){
        int i,u=0,l=strlen(s);
        For(i,0,l-1){int c=idx(s[i]);
            if(!ch[u][c])ch[u][c]=++sz;u=ch[u][c];
        }val[u]=v;}
}tr;
struct AC_machine{int fail[maxn],last[maxn],cnt[maxn];
    inline void init(){memset(fail,0,sizeof(fail));memset(last,0,sizeof(last));memset(cnt,0,sizeof(cnt));}
    inline void getfail(){int c,u,k,r;queue<int>q;fail[0]=0;
        For(c,0,25)if(tr.ch[0][c])q.push(tr.ch[0][c]);
        while(!q.empty()){r=q.front(),q.pop();
            For(c,0,25){u=tr.ch[r][c];
                if(!u){tr.ch[r][c]=tr.ch[fail[r]][c];continue;}q.push(u);
                k=fail[r];while(k&&!tr.ch[k][c])k=fail[k];
                fail[u]=tr.ch[k][c];
                last[u]=tr.val[fail[u]]?fail[u]:last[fail[u]];
            }}}
    inline void calc(int x){while(x)++cnt[tr.val[x]],x=last[x];}
    inline void find(char *s){int i,j,l=strlen(s);
        For(i,0,l-1){j=tr.ch[j][idx(s[i])];
            if(tr.val[j])calc(j); else if(last[j]) calc(last[j]);}}
}ac;
int main(){int i;
    while(scanf("%d",&n),n){tr.init();ac.init();
        For(i,1,n)scanf("%s",P[i]),tr.insert(P[i],i);
        scanf("%s",T);ac.getfail();ac.find(T);
        int best=-1;
        For(i,1,n) if(ac.cnt[i]>best) best=ac.cnt[i];
        printf("%d
",best);
        For(i,1,n)if(ac.cnt[i]==best)printf("%s
",P[i]);
    }return 0;}