HDOJ 2222 AC自动机模板题

Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 44704    Accepted Submission(s): 14109

Problem Description

In the modern time, Search engine came into the life of everybody like Google, Baidu, etc.
Wiskey also wants to bring this feature to his image retrieval system.
Every image have a long description, when users type some keywords to find the image, the system will match the keywords with description of image and show the image which the most keywords be matched.
To simplify the problem, giving you a description of image, and some keywords, you should tell me how many keywords will be match.

 

Input

First line will contain one integer means how many cases will follow by.
Each case will contain two integers N means the number of keywords and N keywords follow. (N <= 10000)
Each keyword will only contains characters 'a'-'z', and the length will be not longer than 50.
The last line is the description, and the length will be not longer than 1000000.

 

Output

Print how many keywords are contained in the description.

 

Sample Input

1
5
she
he
say
shr
her
yasherhs

 

Sample Output

3

 

写了半天，找不到哪里错了，到博客上找了篇文章。 AC自动机

然后加上AC自动机的介绍AC自动机算法详解

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<queue>
using namespace std;
struct node{
       node *fail,*s[27];
       int w;
}*head;
int t,n,i,j,sum;
char temp[51],str[1000001];
queue<node*> myqueue;
node *getfail(node *p,int k){
     if (p->s[k]!=NULL) return p->s[k];
     else 
          if (p==head) return head;
          else return getfail(p->fail,k);
}
void built_trie(){
     node *root=head;
     node *tep;
     for (j=0;j<strlen(temp);j++){
         if (root->s[temp[j]-'a']==NULL){
            tep=new node;
            for (int k=0;k<26;k++)
                tep->s[k]=NULL;
            tep->w=0;
            tep->fail=head;
            root->s[temp[j]-'a']=tep;
            }
         
         root=root->s[temp[j]-'a'];
         if (j==strlen(temp)-1) root->w+=1;
         }
     return ;
}
void built_ac(){
     node *root;
     while (!myqueue.empty()) myqueue.pop();
     myqueue.push(head);
     while (!myqueue.empty()){//构造fail指针用BFS的方法扩展
           root=myqueue.front();
           myqueue.pop();
           for (j=0;j<26;j++)
               if (root->s[j]!=NULL){
                  myqueue.push(root->s[j]);
                  if (root==head) root->s[j]->fail=head;
                  else root->s[j]->fail=getfail(root->fail,j);
                  }
           }
     return ;
}
void find(){
     int len=strlen(str);
     node* tep;
     node *root=head;
     for (j=0;j<len;j++){
         while (root->s[str[j]-'a']==NULL && root!=head) root=root->fail;
         root=(root->s[str[j]-'a']==NULL)?head:root->s[str[j]-'a'];
         tep=root;
         while (tep!=head){//如果单词A出现过了，那么与他具有相同前缀的单词也都出现过了
               sum+=tep->w;
               tep->w=0;
               tep=tep->fail;
               }
         }
     return ;
     }
int main(){
    scanf("%d",&t);
    while (t--){
          sum=0;
          head=new node;
          for (i=0;i<26;i++)
              head->s[i]=NULL;
          head->fail=head;
          head->w=0;
          scanf("%d",&n);
          for (i=0;i<n;i++){
              scanf("%s",temp);
              built_trie();
              }
          built_ac();
          scanf("%s",str);
          find();
          printf("%d
",sum);
          }
    return 0;
}

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