Keywords Search(AC自动机)

柔光的暖阳◎ 2022-08-02 12:56 184阅读 0赞

# Keywords Search #

**Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 43061    Accepted Submission(s): 13550**

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

Author

Wiskey

#include<iostream>
    #include<cstdio>
    #include<cstring>
    
    using namespace std;
    
    const int N=26;
    
    struct Trie{
        Trie *fail;     //失败指针
        Trie *next[N];   //Tire每个节点的26个子节点（最多26个字母）
        int count;       //是否为该单词的最后一个节点
    
        Trie(){     //构造函数初始化
            fail=NULL;
            count=0;
            memset(next,NULL,sizeof(next));
        }
    }*q[500010];    //队列，方便用于bfs构造失败指针
    
    char word[55];  //输入的单词
    char str[1000010];   //模式串
    int head,tail;      //队列的头尾指针
    
    void Insert(char *str,Trie *Root){
        Trie *loc=Root;
        int i=0;
        while(str[i]!='\0'){
            int id=str[i]-'a';
            if(loc->next[id]==NULL)
                loc->next[id]=new Trie();
            loc=loc->next[id];
            i++;
        }
        loc->count++;
    }
    
    void AC_automation(Trie *Root){
        Root->fail=NULL;
        q[head++]=Root;
        Trie *cur,*tmp;
        while(head!=tail){
            cur=q[tail++];
            tmp=NULL;
            for(int i=0;i<26;i++){
                if(cur->next[i]!=NULL){
                    if(cur==Root)
                        cur->next[i]->fail=Root;
                    else{
                        tmp=cur->fail;
                        while(tmp!=NULL){
                            if(tmp->next[i]!=NULL){
                                cur->next[i]->fail=tmp->next[i];
                                break;
                            }
                            tmp=tmp->fail;
                        }
                        if(tmp==NULL)
                            cur->next[i]->fail=Root;
                    }
                    q[head++]=cur->next[i];
                }
            }
        }
    }
    
    int query(Trie *Root){
        int i=0,cnt=0;
        Trie *loc=Root,*tmp;
        while(str[i]!='\0'){
            int id=str[i]-'a';
            while(loc->next[id]==NULL && loc!=Root)
                loc=loc->fail;
            loc=loc->next[id];
            loc=(loc==NULL)?Root:loc;
            tmp=loc;
            while(tmp!=Root&&tmp->count!=-1){//注意这里只是计算字串的个数，相同的算一个！
                cnt+=tmp->count;//这个就是通过这里实现的！比如字串a父串为aaa应该输入1而不是3
                tmp->count=-1;
                tmp=tmp->fail;
            }
            i++;
        }
        return cnt;
    }
    /*
    abc
    abcabcabc
    */
    int main(){
    
        //freopen("input.txt","r",stdin);
    
        int t,n;
        scanf("%d",&t);
        while(t--){
            head=tail=0;
            Trie *Root=new Trie();
            scanf("%d",&n);
            getchar();
            while(n--){
                gets(word);
                Insert(word,Root);
            }
            AC_automation(Root);
            scanf("%s",str);
            printf("%d\n",query(Root));
        }
        return 0;
    }