HDU2222(AC自动机入门题)

Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 48520    Accepted Submission(s): 15473

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

 

 

学习AC自动机的博客：

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

 http://www.cnblogs.com/kuangbin/p/3164106.html

学习kuangbin大牛的模板。

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int N=;
const int MAXN=;
struct Trie{
    int next[MAXN][N],fail[MAXN],end[MAXN];
    int root;
    int tot;
    int newnode()
    {
        for(int i=;i<N;i++)
                next[tot][i]=-;
        end[tot++]=;
        return tot-;
    }
    void init()
    {
        tot=;
        root=newnode();
    }

    void insert(char buf[])
    {
        int len=strlen(buf);
        int now=root;
        for(int i=;i<len;i++)
        {
            int k=buf[i]-'a';
            if(next[now][k]==-)
                next[now][k]=newnode();
            now=next[now][k];
        }
        end[now]++;
    }

    void build()
    {
        queue<int> que;
        fail[root]=root;
        for(int i=;i<N;i++)
            if(next[root][i]==-)
                next[root][i]=root;
            else
            {
                fail[next[root][i]]=root;
                que.push(next[root][i]);
            }

        while(!que.empty())
        {
            int now = que.front();
            que.pop();
            for(int i=;i<N;i++)
                if(next[now][i]==-)
                    next[now][i]=next[fail[now]][i];
                else
                {
                    fail[next[now][i]]=next[fail[now]][i];
                    que.push(next[now][i]);
                }
        }
    }

    int query(char buf[])
    {
        int len=strlen(buf);
        int now=root;
        int res=;
        for(int i=;i<len;i++)
        {
            now=next[now][buf[i]-'a'];
            int temp=now;
            while(temp!=root)
            {
                res+=end[temp];
                end[temp]=;//模式串只在主串中匹配一次就可以了
                temp=fail[temp];
            }
        }
        return res;
    }

};
Trie ac;
char buf[MAXN+MAXN];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n;
        scanf("%d",&n);
        ac.init();
        for(int i=;i<n;i++)
        {
            scanf("%s",buf);
            ac.insert(buf);
        }
        ac.build();
        scanf("%s",buf);
        printf("%d\n",ac.query(buf));
    }

    return ;
}