hdu 2222 Keywords Search (AC自动机)

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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，通过构建自动机并利用失配指针来高效查找文本中出现的关键词，确保关键词不重复匹配。

Keywords Search

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 71488 Accepted Submission(s): 24396

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

Recommend

lcy

题意：

给出了N个子串问在主串中能匹配多少个子串。子串不能重复出现在主串中。

思路：

用AC自动机，cnt记录下到达该节点的完整的子串的个数。然后用失配指针找到所有的可以匹配到的，在计算了以这个结点为结束的子串后要把cnt赋值为-1，防止它再次计算。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int maxn=1000005;
char str[maxn];
struct node
{
    node *next[26];
    node *fail;
    int cnt;
    node()
    {
        memset(next,NULL,sizeof(next));
        fail=NULL;
        cnt=0;
    }
};
queue<node *> que;
void in(node *root,char str[])
{
    node *now=root;
    int i=0;
    while(str[i])
    {
        int x=str[i]-'a';
        if(now->next[x]==NULL)
        {
            now->next[x]=new node();
        }
        now=now->next[x];
        i++;
    }
    now->cnt++;
}
void makefail(node *root)
{
    root->fail=NULL;
    que.push(root);
    while(!que.empty())
    {
        node *now=que.front();
        que.pop();
        node *p=NULL;
        for(int i=0;i<26;i++)
        {
            if(now->next[i]!=NULL)
            {
                if(now==root) now->next[i]->fail=root;
                else
                {
                    p=now->fail;
                    while(p!=NULL)
                    {
                        if(p->next[i]!=NULL)
                        {
                            now->next[i]->fail=p->next[i];
                            break;
                        }
                        p=p->fail;
                    }
                    if(p==NULL) now->next[i]->fail=root;
                }
                que.push(now->next[i]);
            }
        }
    }
}
int query(node *root,char str[])
{
    node* now=root;
    int ans=0;
    int len=strlen(str);
    for(int i=0;i<len;i++)
    {
        int x=str[i]-'a';
        while(now->next[x]==NULL&&now!=root) now=now->fail;
        now=now->next[x];
        if(now==NULL)
            now=root;
        node* tmp=now;
        while(tmp!=NULL&&tmp->cnt!=-1)
        {
            ans+=tmp->cnt;
            tmp->cnt=-1;
            tmp=tmp->fail;
        }
    }
    return ans;
}
int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        node *root=new node();
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%s",str);
            in(root,str);
        }
        makefail(root);
        scanf("%s",str);
        printf("%d\n",query(root,str));
    }
    return 0;
}