hdu 2222 ac自动机入门题 可以做模板

Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25141    Accepted Submission(s): 8224

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

 

Author

Wiskey

 

Recommend

lcy

 

 

http://acm.hdu.edu.cn/showproblem.php?pid=2222

 

 

建议不了解ac自动机的 请看看

http://hi.baidu.com/nialv7/item/ce1ce015d44a6ba7feded52d

http://www.cppblog.com/mythit/archive/2009/04/21/80633.html

 

题意：

题意：给出n个串，然后给一篇文章，问这n个串有多少个在文章里面出现过。。。

trick：n个串可能有相同的，需按照不同串处理。

 

#include
#include
#include
#include
using namespace std;
char str[1000000+100];

struct node
{
    int count;
    struct node *next[26];
    struct node *fail;
    void init()
    {
        int i;
        for(i=0;i<26;i++)
            next[i]=NULL;
        count=0;
        fail=NULL;
    }
}*root;
void insert()
{
    int len,k;
    node *p=root;
    len=strlen(str);
    for(k=0;knext[pos]==NULL)
        {
            p->next[pos]=new node;
            p->next[pos]->init();
            p=p->next[pos];
        }
        else 
            p=p->next[pos];
    }
    p->count++;
}
void getfail()
{
    int i;
       node *p=root,*son,*temp;
       queueque;
       que.push(p); 
       while(!que.empty())
       {
           temp=que.front();
           que.pop();
           for(i=0;i<26;i++)
           {
               son=temp->next[i];
               if(son!=NULL)
               {
                   if(temp==root) {son->fail=root;}
                   else
                   {
                       p=temp->fail;
                       while(p)
                       {
                           if(p->next[i])
                           {
                               son->fail=p->next[i];
                               break;
                           }
                           p=p->fail;
                       }
                       if(!p)  son->fail=root;
                   }
                   que.push(son);
               }
           }
       }
}
void query()
{
    int len,i,cnt=0;
    len=strlen(str);
    node *p,*temp;
    p=root;
    for(i=0;inext[pos]&&p!=root)  p=p->fail;      
        p=p->next[pos];//
        if(!p) p=root;//
        temp=p;
        /*不要用*temp=*p  因为*p表示一个node，而*temp也表示一个node 但是由于*temp没有分配空间 所以是不能进行赋值的 但是可以用temp指针去指向p*/
        while(temp!=root)
        {
            if(temp->count>=0) 
            {
                cnt+=temp->count;
                temp->count=-1;  
            }
            else break; 
            temp=temp->fail; 
        }
    }
    printf("%d\n",cnt);
}
int main()
{
    int cas,n;
    scanf("%d",&cas);
    while(cas--)
    {
        root=new node;
        root->init();
        root->fail=NULL;
        scanf("%d",&n);
        int i;
        getchar();
        for(i=0;i