hdu2222之AC自动机入门 - c++编程基础

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hdu2222之AC自动机入门

2014-11-23 22:13:32 【大中小】浏览:8860次

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 retrieva l 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   
#include   
#include   
#include   
#include   
#include   
#include   
#include   
#include   
#define INF 9999999   
using namespace std;  
  
const int MAX=1000000+10;  
char a[60],b[MAX];  
  
struct TrieNode{  
    int num;//记录单词的数量   
    TrieNode *next[26],*fail;//下一个节点和失败指针   
    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;  
    queueq;  
    q.push(p);  
    while(!q.empty()){//一层一层的构造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来判断是否已经寻找过    
            sum+=next->num;  
            next->num=-1;//这里注意一定要赋值为负数,不能为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>a;  
            InsertNode(a);//插入单词    
        }  
        Build_AC();//构造失败指针,也就是建立AC自动机的精髓   
        cin>>b;  
        cout<


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