hdu 2222 Keywords Search ac自己主动机

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Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 42838    Accepted Submission(s): 13488

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

 

给出n个字符串和一个模式串 问在模式串中出现了多少个给出的字符串

ac自己主动机模板题

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
char str1[55],str2[1111111];
struct node{
    node *fail,*next[26];
    int ed;
    node(){
        ed=0;
        fail=NULL;
        for(int i=0;i<26;i++)
            next[i]=NULL;
    }
}*q[511111],*root;
void tree_insert(char *str){
    node *p=root;
    int l=strlen(str);
    int id;
    for(int i=0;i<l;i++){
        id=str[i]-'a';
        if(p->next[id]==NULL){
            p->next[id]=new node();
        }
        p=p->next[id];
    }
    p->ed++;
}
void build_ac_automachine(){
    int i;
    node *temp,*p;
    root->fail=NULL;
    int fr,ed;
    fr=ed=0;
    q[ed++]=root;
    while(fr<ed){
        temp=q[fr++];
        for(i=0;i<26;i++){
            if(temp->next[i]!=NULL){
                if(temp==root)
                    temp->next[i]->fail=root;
                else {
                    p=temp->fail;
                    while(p){
                        if(p->next[i]){
                            temp->next[i]->fail=p->next[i];
                            break;
                        }
                        p=p->fail;
                    }
                    if(p==NULL)
                        temp->next[i]->fail=root;
                }
                q[ed++]=temp->next[i];
            }
        }
    }
}
int query(char *str){
    int cnt=0;
    int l=strlen(str);
    node *p=root;
    for(int i=0;i<l;i++){
        int id=str[i]-'a';
        while(p->next[id]==NULL&&p!=root)
            p=p->fail;
        p=p->next[id];
        if(p==NULL)
            p=root;
        node *temp=p;
        while(temp!=root&&temp->ed!=-1){
            cnt+=temp->ed;
            temp->ed=-1;
            temp=temp->fail;
        }
    }
    return cnt;
}
int main(){
    int t,n;
    scanf("%d",&t);
    while(t--){
        root=new node();
        scanf("%d",&n);
        while(n--){
            scanf("%s",str1);
            tree_insert(str1);
        }
        scanf("%s",str2);
        build_ac_automachine();
        printf("%d
",query(str2));
    }
    return 0;
}