hdu 2222 Keywords Search ac自动机模板

Keywords Search

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 33671 Accepted Submission(s): 10899



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

Author

Wiskey



Recommend

lcy



ac自动机，指针版代码

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
using namespace std;
const int N = 55005;
int cnt = 0;
struct trie{
    trie *next[26],*fail,*last;
    int  id;
    int cnt;
    trie(){
        for(int i = 0; i < 26; ++i) next[i] = NULL;
        fail = last = 0;
        id = 0;
        cnt = 0;
    }
} *e[N * 6];
void init()
{
    for(int i = 0; i < cnt; i++){
           delete e[i];
    }
      cnt = 0;
}
class AAM{
   public:
    trie *root;
    void init(){
        root = e[cnt++] = new trie();
    }
    void insert(char *s){
        int len = strlen(s);
        trie * p = root;
        for(int i = 0; i < len; ++i){
            int index = s[i] - 'a';
            if(p -> next[index] == NULL) p -> next[index] = e[cnt++] = new trie();
            p = p -> next[index];
        }
        p -> id ++;
    }
   void getfail(){
        queue<trie *>Q;
        trie *p;
        root -> last = root -> fail = root;
        for(int i = 0; i < 26; ++i){
            if(!root -> next[i]) root -> next[i] = root;
            else Q.push(root -> next[i]);
            root -> next[i] -> fail = root;
            root -> next[i] -> last = root;

        }
        while(!Q.empty()){
            p = Q.front(); Q.pop();
            for(int i = 0; i < 26; ++i){
                    if(!p -> next[i]){
                            p -> next[i] = p -> fail -> next[i];

                    }else{
                            p -> next[i] -> fail = p -> fail -> next[i];
                            p -> next[i] -> last = (p -> next[i] ->fail -> id) ?
                            p -> next[i] -> fail : p -> next[i] -> fail -> last;

                            Q.push(p -> next[i]);
                    }
            }
        }
   }
   int num;
   void count(trie *p)
   {
       if(p == root) return;
       if(p -> id && p -> cnt == 0){
            num += p -> id; p -> cnt = 1;
       }
                count(p -> last);
   }
   int query(char *s){
            num = 0;
            trie *p = root;
            int len = strlen(s);
            for(int i =  0; i < len; ++i){
                    int index = s[i] - 'a';
                    p = p -> next[index];
                    if(p -> id) count(p);
                    else count(p -> last);
            }
            return num;
   }
};
char s[10005];
int main()
{
    int t,n;
    AAM T;
    scanf("%d",&t);
    while(t--){
        init();T.init();
        scanf("%d",&n);
        while(n--) scanf("%s",s),T.insert(s);
        T.getfail();
        scanf("%s",s);
        printf("%d\n",T.query(s));
    }
    return 0;
}

数组版

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
using namespace std;
const int N = 300000;
int ch
[26];
int fail
,last
,val
,vis
;
class AAM{
    int cnt,root;
public:
    void init(){
       cnt = 0, root = cnt++;
       memset(vis,0,sizeof(vis));
       memset(ch[0],0,sizeof(ch[0]));
       memset(val,0,sizeof(val));
    }
    void insert(char *s){
            int len = strlen(s);
            int p = root;
            for(int i = 0; i < len; ++i){
                    int index = s[i] - 'a';
                    if(!ch[p][index]){
                            memset(ch[cnt],0,sizeof(ch[cnt]));
                            ch[p][index] = cnt++;
                    }
                    p = ch[p][index];
            }
            val[p]++;
    }
    void getfail(){
            int p = root;
            queue<int>Q;
            fail[p] = last[p] = p;
            for(int i = 0;i < 26; ++i){
                int u = ch[p][i];
                if(u) {last[u] = fail[u] = p;Q.push(u);}
                else ch[p][i] = p;
            }
            while(!Q.empty()){
                int p = Q.front(); Q.pop();
                for(int i = 0; i < 26; ++i){
                        int &u = ch[p][i];
                        if(u) {
                                fail[u] = ch[fail[p]][i];
                                last[u] = val[fail[u]]?fail[u]:last[fail[u]];
                                Q.push(u);
                        }else u = ch[fail[p]][i];
                }
            }
    }

    int num;
    void count(int p){
        if(p == root) return;
        if(!vis[p] && val[p])num += val[p],vis[p] = 1;
        count(last[p]);
    }
    int query(char *s){
        num = 0;
        int len = strlen(s);
        int p =  root;
        for(int i = 0; i < len ; ++i){
            int index = s[i] - 'a';
            p = ch[p][index];
            count(p);
        }
        return num;
    }
};
char s[1000000];

int main(){

    int t,n;
    AAM T;
    scanf("%d",&t);
    while(t--){
        T.init();
        scanf("%d",&n);
        while(n--) scanf("%s",s),T.insert(s);
        T.getfail();
        scanf("%s",s);
        printf("%d\n",T.query(s));
    }

}