hdu2222之AC自動機入門

Keywords Search
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25799    Accepted Submission(s): 8421

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

#include<iostream>   
#include<cstdio>   
#include<cstdlib>   
#include<cstring>   
#include<string>   
#include<queue>   
#include<algorithm>   
#include<map>   
#include<iomanip>   
#define INF 9999999   
using namespace std;  
  
const int MAX=1000000+10;  
char a[60],b[MAX];  
  
struct TrieNode{  
    int num;//記錄單詞的數量   
    TrieNode *next[26],*fail;//下一個節點和失敗指針   
    TrieNode(){  
        num=0;  
        fail=0;  
        memset(next,0,sizeof next);  
    }   
}*root;  
  
void InsertNode(char *a){  
    int k=0;  
    TrieNode *p=root;  
    while(a[k]){  
        if(!p->next[a[k]-'a'])p->next[a[k]-'a']=new TrieNode;  
        p=p->next[a[k++]-'a'];  
    }  
    ++p->num;  
}  
  
void Build_AC(){  
    int k=0;  
    TrieNode *p=root,*next;  
    queue<TrieNode *>q;  
    q.push(p);  
    while(!q.empty()){//一層一層的構造fail    
        p=q.front();  
        q.pop();  
        for(int i=0;i<26;++i){  
            if(p->next[i]){  
                next=p->fail;  
                while(next){  
                    if(next->next[i]){p->next[i]->fail=next->next[i];break;}  
                    next=next->fail;  
                }  
                if(!next)p->next[i]->fail=root;  
                q.push(p->next[i]);  
            }  
        }  
    }  
}  
  
int SearchTrie(char *a){  
    int k=0,sum=0;  
    TrieNode *p=root,*next;   
    while(a[k]){  
        while(!p->next[a[k]-'a'] && p != root)p=p->fail;  
        p=p->next[a[k++]-'a'];  
        if(!p)p=root;  
        next=p;  
        while(next != root && next->num != -1){//不能用0來判斷是否已經尋找過    
            sum+=next->num;  
            next->num=-1;//這裡注意一定要賦值為負數,不能為0    
            next=next->fail;  
        }  
    }  
    return sum;  
}  
  
void Free(TrieNode *p){  
    for(int i=0;i<26;++i)if(p->next[i])Free(p->next[i]);  
    delete p;  
}  
  
int main(){  
    int T,n;  
    cin>>T;  
    while(T--){  
        root=new TrieNode;  
        cin>>n;  
        for(int i=0;i<n;++i){  
            cin>>a;  
            InsertNode(a);//插入單詞    
        }  
        Build_AC();//構造失敗指針,也就是建立AC自動機的精髓   
        cin>>b;  
        cout<<SearchTrie(b)<<endl;//查詢含有的單詞數量   
        Free(root);  
    }  
    return 0;  
}  
/*  
10 
2 
abcdef 
bcd 
abcdef 
1 
h 
hhhhh 
5 
bhea 
her 
he 
h 
ha 
bhera 
5 
bhea 
her 
he 
h 
ha 
bhera 
*/