sdut trie

Trie

Description

Encounter the word do not know how to do? Dictionary ah, there are n known dictionary words, assuming that words are in lowercase letters. M existing words do not know, ask the m word appears in the dictionary.

Input

Contain multiple sets of test cases.

The first input line n, m (n> = 0 && n <= 100000 && m> = 0 && m <= 100000) are present in the dictionary and n words of m words to the query.

Followed by n lines, represents the presence of the word in the dictionary.

Then the line m, m to query words

n = 0 && m = 0 the routine ends

To ensure that all data words are all lowercase letters, and a length of not more than 10

Output

If there is output Yes, there is no output No.

Sample

Input 

3 2 aab aa ad ac ad 0 0

 

 

 

 

 

 

Output

No Yes

 

 

 

 

Such as free practice and then add the list

#include <bits / STDC ++ H.> the using namespace STD;
 char S [ 20 is ];
 int Trie [ 500.01 thousand ] [ 26 is ];         // array is open metaphysical, bigger TLE smaller RTE int CC [ 500.01 thousand ];         // It used to prove the word exists int F = . 1 ; void in_sert () {
     int K = strlen (S);
     int P = 0 ;
     for ( int I = 0 ; I <K; I ++ ) {
         int C = S [I] - ' A

 



' ;
         IF (! Trie [the p-] [c]) { 
            memset (Trie [f], 0 , sizeof (Trie [f]));         // can be used to re-initialize the time, should be able to save some time? 
            Trie [p] [C] = F; 
            F ++ ; 
        } 
        p = Trie [p] [C];     // next letter p points to a storage location 
    } 
    CC [p] = . 1 ;     // tag word at the ending 
} 

BOOL se_arch () {
     int K = strlen (S);
     int P = 0 ;
     for ( int= I 0 ; I <K; I ++ ) {
         int C = S [I] - ' A ' ;
         IF (Trie [P] [C] == 0 ) return  0 ;     // described next letter is not a tree, the end of 
        = P Trie [P] [C]; 
    } 
    IF (CC [P]!) return  0 ;         // in the tree is not a complete word, for example, AAA tree saved, the query AA 
    the else  return  . 1 ; 
} 

int main () 
{ 
    int n-, m;
     the while (~ Scanf ( " % D% D " , & n-, & m) && (n-||m)) { 
        getchar (); 
        Memset (Trie [ 0 ], 0 , the sizeof (Trie [ 0 ])); 
        Memset (CC, 0 , the sizeof (CC)); 
        F = . 1 ;     // for indicating letters stored the position of 
        the while (N-- ) { 
            Scanf ( " % S " , S); 
            in_sert (); 
        } 
        the while (M-- ) { 
            Scanf ( " % S " , S);
             int T = se_arch ();
            if(t) printf("Yes\n");
            else printf("No\n");
        }
    }
    return 0;
}

 

 

#include <bits / STDC ++ H.>
the using namespace STD; char S [20 is]; int Trie [500.01 thousand] [26 is]; // open array is metaphysical, bigger TLE smaller RTEint cc [500010]; // with to prove the word exists int F =. 1;
void in_sert () {int K = strlen (S); int P = 0; for (int I = 0; I <K; I ++) {int C = S [I] - ' a '; if (trie [p ] [c]!) {memset (trie [f], 0, sizeof (trie [f])); // can be used to re-initialize the time, should be able to save some time? trie [p] [c] = f; f ++;} p = trie [p] [c]; // p points to the storage location of a letter} cc [p] = 1; // tag word at the ending}
BOOL se_arch () {int k = strlen ( s); int p = 0; for (int i = 0; i <k; i ++) {int c = s [i] - 'a'; if (trie [p] [c ] == 0) return 0; // not be described next letter in the tree, the end of p = trie [p] [c ];} if (cc [p]) return 0;! // in the tree is not a complete word , such as trees stored AAA, query aa else return 1;}
int main(){    int n, m;    while(~scanf("%d%d", &n, &m)&&(n||m)){        getchar();        memset(trie[0], 0, sizeof(trie[0]));        memset(cc, 0, sizeof(cc));        f = 1;//用来指示字母保存的位置        while(n--){            scanf("%s", s);            in_sert();        }        while(m--){            scanf("%s", s);            int t = se_arch();            if(t) printf("Yes\n");            else printf("No\n");        }    }    return 0;}