HDU 2222 Keywords Search AC自动机入门模版题

最新推荐文章于 2020-06-23 16:31:38 发布

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本文介绍了一种使用AC自动机实现关键词搜索的方法。通过构建AC自动机，可以在文本中高效地查找多个关键词出现的次数。文章提供了完整的代码示例，并解释了如何插入关键词、建立AC自动机以及进行查询。

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Keywords Search

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 30691 Accepted Submission(s): 9999

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

Author

Wiskey

给你n个单词，然后一个文本串，让你在这个文本串里面找出现了几个单词。
AC自动机入门，详解：点击打开链接

//234MS	29364K
#include<stdio.h>
#include<string.h>
char str[1000007];
int head,tail;//队列的头和尾
struct node
{
    node *next[26];//Trie每个节点的26个字节点
    node *fail;//失效指针
    int count;//是否为该单词的最后一个节点
    node()//构造函数初始化
    {
        count=0;
        fail=NULL;
        memset(next,0,sizeof(next));
    }
} *q[500001];//队列，方便用于bfs构造失效指针
void insert(node *root)
{
    node *p=root;
    int i=0,index;
    while(str[i])
    {
        index=str[i]-'a';
        if(p->next[index]==NULL)p->next[index]=new node();
        p=p->next[index];
        i++;
    }
    p->count++;//在单词的最后一个节点count+1，代表一个单词
}
void build_ac(node *root)
{
    root->fail=NULL;
    q[head++]=root;
    while(head!=tail)
    {
        node *temp=q[tail++];
        node *p=NULL;
        for(int i=0;i<26;i++)
        {
            if(temp->next[i]!=NULL)
            {
                if(temp==root)temp->next[i]->fail=root;
                else
                {
                    p=temp->fail;
                    while(p!=NULL)
                    {
                        if(p->next[i]!=NULL)
                        {
                            temp->next[i]->fail=p->next[i];
                            break;
                        }
                        p=p->fail;
                    }
                    if(p==NULL)temp->next[i]->fail=root;
                }
                q[head++]=temp->next[i];
            }
        }
    }
}
int query(node *root)
{
    int i=0,cnt=0,index;
    node *p=root;
    while(str[i])
    {
        index=str[i]-'a';
        while(p->next[index]==NULL&&p!=root)p=p->fail;
        p=p->next[index];
        p=(p==NULL)?root:p;
        node *temp=p;
        while(temp!=root&&temp->count!=-1)
        {
            cnt+=temp->count;
            temp->count=-1;
            temp=temp->fail;
        }
        i++;
    }
    return cnt;
}
int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        head=tail=0;
        node *root=new node();
        scanf("%d",&n);
        while(n--)
        {
            scanf("%s",str);
            insert(root);
        }
        build_ac(root);
        scanf("%s",str);
        printf("%d\n",query(root));
    }
    return 0;
}