HDU-2222 Keywords Search AC自动机

Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10744    Accepted Submission(s): 3722


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
 
　　该题为AC自动机模板题，果然它还是很强大的，该题要注意单词有重复出现的。
　　代码如下：
  1 #include <cstring>
  2 #include <cstdio>
  3 #include <cstdlib>
  4 #include <queue>
  5 using namespace std;
  6 
  7 struct Node
  8 {
  9     int end;
 10     Node *fail, *child[26];
 11 };
 12 
 13 char s[1000005];
 14 
 15 void getstr( char *s )
 16 {
 17     char c;
 18     int p = 0;
 19     while( c = getchar(), c == ' ' || c == '\n' ) ;
 20     s[p++] = c;
 21     while( c = getchar(), c != ' ' && c != '\n' )
 22         s[p++] = c;
 23     s[p] = '\0';
 24 }
 25 
 26 Node *init(  )
 27 {
 28     Node *r = ( Node * )malloc( sizeof( Node ) );
 29     r->end = 0;
 30     r->fail = NULL;
 31     memset( r->child, NULL, sizeof( r->child ) );
 32     return r;
 33 }
 34 
 35 void creattree( Node *p, char *in )
 36 {
 37     int index = *in - 'a';
 38     if( p->child[index] == NULL )
 39         p->child[index] = init();
 40     if( *( in + 1 ) )
 41         creattree( p->child[index], in + 1 );
 42     else
 43         p->child[index]->end++;
 44 }
 45 
 46 void creatfail( Node *p )
 47 {
 48     queue< Node * >q;
 49     q.push( p );
 50     while( !q.empty() )
 51     {
 52         Node *pos = q.front();
 53         q.pop();
 54         for( int i = 0; i < 26; ++i )
 55         {
 56             Node *f = pos->fail;
 57             if( pos->child[i] )
 58             {
 59                 while( f != NULL )
 60                 {
 61                     if( f->child[i] )
 62                     {
 63                         pos->child[i]->fail = f->child[i];
 64                         break;
 65                     }
 66                     else
 67                         f = f->fail;
 68                 }
 69                 if( f == NULL )
 70                 {
 71                     pos->child[i]->fail = p;
 72                 }
 73                 q.push( pos->child[i] );
 74             }
 75         }
 76     }
 77 }
 78 
 79 int acauto( Node *p, char *text )
 80 {
 81     int ans = 0, lenth = 0, len = strlen( text );
 82     Node *f = p;
 83     for( int i = 0; i < len; ++i )
 84     {
 85         int type = text[i] - 'a';
 86         while( f )
 87         {
 88             if( f->child[type] )
 89             {
 90                 f = f->child[type];
 91                 break;
 92             }
 93             else
 94                 f = f->fail;
 95         }
 96 
 97         if( f == NULL ) // 如果都已经找到了根节点仍不匹配
 98         {
 99             f = p; // 从头开始
100             continue;
101         }
102 
103         Node *t = f;
104         while( t != p ) // 成功匹配一个字符的更新操作
105         {
106             if( t->end )
107             {
108                 ans += t->end;
109                 t->end = 0;
110             }
111             t = t->fail;
112         }
113     }
114     return ans;
115 }
116 
117 void _free( Node *p )
118 {
119     for( int i = 0; i < 26; ++i )
120     {
121         if( p->child[i] )
122             _free( p->child[i] );
123     }
124     free( p );
125 }
126 
127 int main()
128 {
129     int T;
130     scanf( "%d\n", &T );
131     while( T-- )
132     {
133         int N;
134         Node *p = init();
135         scanf( "%d", &N );
136         for( int i = 0; i < N; ++i )
137         {
138             getstr( s );
139             creattree( p, s );
140         }
141         getstr( s );
142         creatfail( p );
143         printf( "%d\n", acauto( p, s ) );
144     }
145 }

　　

posted @ 2011-09-04 03:04  沐阳  阅读(848)  评论(0)  编辑  收藏  举报