Keywords Search （HDU-2222）（AC自动机模板题）

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

题意：给你一个长度为n的单词表，一个文本串，问你这个文本串中出现了单词表中多少个单词。

思路：直接套用AC自动机的模板即可。

AC代码：

#include 
typedef long long ll;
const int maxx=500010;
const int maxn=26;
const int inf=0x3f3f3f3f;
using namespace std;
int Trie[maxx][maxn];
int fail[maxx];
int cntword[maxx];
//bool visit[maxx];//另一种模板
char s[1000010];
int k;
void init()
{
    memset(Trie,0,sizeof(Trie));
    memset(cntword,0,sizeof(cntword));
    //memset(visit,0,sizeof(visit));
    memset(fail,0,sizeof(fail));
    k=0;
}
void insert(char *s)
{
    int p=0;
    int len=strlen(s);
    for(int i=0; iq;
    for(int i=0; i<26; i++)
    {
        if(Trie[0][i])
        {
            fail[Trie[0][i]]=0;
            q.push(Trie[0][i]);
        }
    }
    while(!q.empty())
    {
        int now=q.front();
        q.pop();
        for(int i=0; i<26; i++)
        {
            if(Trie[now][i])
            {
                fail[Trie[now][i]]=Trie[fail[now]][i];
                q.push(Trie[now][i]);
            }
            else
                Trie[now][i]=Trie[fail[now]][i];
        }
    }
}
int query(char *s)
{
    int now=0;
    int ans=0;
    int len=strlen(s);
//    for(int i=0; i