Keywords Search - HDU 2222 AC自动机

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本文介绍了一种使用AC自动机进行高效关键词匹配的方法。通过构建AC自动机，可以在描述文本中快速查找并计数所有关键词出现的次数。文章提供了一个具体的实现案例，包括构建AC自动机、插入关键词和查询关键词等步骤。

Keywords Search

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 36244 Accepted Submission(s): 11733

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

题意：对于一个字符串，存在多少个关键字。

思路：AC自动机。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
int next[500010][26],fail[500010],end[500010],L;
char s[1000010];
int newnode()
{
    int i;
    for(i=0;i<26;i++)
      next[L][i]=-1;
    end[L++]=0;
    return L-1;
}
void Insert()
{
    int now=0,len=strlen(s+1),i;
    for(i=1;i<=len;i++)
    {
        if(next[now][s[i]-'a']==-1)
          next[now][s[i]-'a']=newnode();
        now=next[now][s[i]-'a'];
    }
    end[now]++;
}
void build()
{
    queue<int> qu;
    fail[0]=0;
    int i;
    for(i=0;i<26;i++)
       if(next[0][i]==-1)
         next[0][i]=0;
       else
       {
           fail[next[0][i]]=0;
           qu.push(next[0][i]);
       }
    while(!qu.empty())
    {
        int now=qu.front();
        qu.pop();
        for(i=0;i<26;i++)
           if(next[now][i]==-1)
             next[now][i]=next[fail[now]][i];
           else
           {
               fail[next[now][i]]=next[fail[now]][i];
               qu.push(next[now][i]);
           }
    }
}
int query()
{
    int i,now=0,temp,ret=0,len=strlen(s+1);
    for(i=1;i<=len;i++)
    {
        now=next[now][s[i]-'a'];
        temp=now;
        while(temp)
        {
            ret+=end[temp];
            end[temp]=0;
            temp=fail[temp];
        }
    }
    return ret;
}
int main()
{
    int T,t,n,i,j,k,ans;
    scanf("%d",&T);
    for(t=1;t<=T;t++)
    {
        scanf("%d",&n);
        L=0;
        newnode();
        for(i=1;i<=n;i++)
        {
            scanf("%s",s+1);
            Insert();
        }
        build();
        scanf("%s",s+1);
        ans=query();
        printf("%d\n",ans);
    }
}