AC自动机（Aho-Corasick Automaton Algorithm)

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AC自动机（Aho-Corasick Automaton Algorithm)

(1)建trie树

(2)广度优先搜索构建所有结点的fail

构造next指针，概括起来是：假如当前的节点的字符为e，沿着它的父亲的next指针走，直到找到一个结点，它的孩子也有e这个字符,然后将当前结点的next指针指向那个结点
有e这个孩子的孩子e。

从root开始广度优先搜索每个点，如果当前搜的点的孩子，与该点的next指针指向的点有相同的孩子，则当前点的孩子指向next指针的孩子，

否则next指针循环搜回去，如果搜不到就指向root。

（3）查找
当时的文本字符串为text,长度为len
则有：
ptr=root;
for i:=1 to len-1 do
    while(ptr->text[i]==NULL&&ptr!=root)
         ptr=ptr->next;
    ptr=ptr->child[i];
   if(ptr==NULL) then ptr=root;//如果搜不到回到根结点
HDOJ2002
Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8434    Accepted Submission(s): 2828

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

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
struct trie
{
    trie* next;
    int  isend;
    trie* child[26];
    trie()
    {
        isend=0;
        memset(child,NULL,sizeof(child));
        next=NULL;
        
    }

}*root;
void insert(char* str)
{
    int num;
    int len;
    trie* ptr;
    ptr=root;
    int i;
    len=strlen(str);
    for(i=0;i<len;i++)
    {
        num=str[i]-'a';
        if(ptr->child[num]==NULL)
        {
            ptr->child[num]=new trie();
        }
        ptr=ptr->child[num];
        
    }
    ptr->isend++;//lable the end of string
}
void ac_automation()
{
    int i;
   queue<trie*> qu;
   trie* current=NULL;
   qu.push(root);
   trie* fail=NULL;
   
   while(!qu.empty())
   {
        current=qu.front();
        qu.pop();
    /*    if(¤t==&0x00370c20)
        {
            int debug=1;
        }*/
    
        //if(current->isend!=1)
        /*    if(current==root)
            {
                for(i=0;i<26;i++)
                {
                    current->child[i]->next=root;
                    qu.push(current->child[i]);

                continue;
            }*/
            
            for(i=0;i<26;i++)
            {
                if(current->child[i]!=NULL)
                {
                    fail=current->next;
                    if(current==root)
                    {
                        current->child[i]->next=root;
                        qu.push(current->child[i]);
                        continue;
                    }
                    while(1)
                    {
                        if(fail==NULL)
                        {
                            current->child[i]->next=root;
                            break;
                        }
                        else
                        {
                            if(fail->child[i]!=NULL)
                            {
			      current->child[i]->next=fail->child[i];//have searched the character of current->child[i]
			      break;
                            }
                        }
                        fail=fail->next;//search the character of current->child[i] follow the next pointer
                    }
                    qu.push(current->child[i]);//push the child of current into queue
                }
            }
        }

        


   
}
char text[1000003];
int ans;
void query()
{
    int len;
    len=strlen(text);
    trie* ptr;
    ptr=root;
    int num;
    int i;
    //trie* fail=root;
    trie* temp;
    for(i=0;i<len;i++)
    {
        num=text[i]-'a';
        while(ptr->child[num]==NULL&&ptr!=root)
        {
            ptr=ptr->next;
        }
        ptr=ptr->child[num];
        if(ptr==NULL)
        {
            ptr=root;
        }
        temp=ptr;
        while(temp!=root&&temp->isend!=-1)//search the matching string follow the next pointer
        {

	  ans+=temp->isend;
            temp->isend=-1;
            temp=temp->next;

        }
        
        
        




    }
}
int main()
{
    int t;
    int n;
    int i;
    char str[52];
    
    scanf("%d",&t);
    while(t--)
    {
        root=new trie();

        scanf("%d",&n);

        for(i=0;i<n;i++)
        {
            scanf("%s",str);
            insert(str);
        }
        ac_automation();
        scanf("%s",text);
        ans=0;
        //len=strlen(text);
        query();
        printf("%d/n",ans);
        
        

    }
    return 0;
}