coreBugZJ

此 blog 已弃。

Keywords Search,HDOJ 2222

Keywords Search

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

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


AC 自动机

我的代码:

  1 #include <iostream>
  2 #include <cstdio>
  3 
  4 using namespace std;
  5 
  6 const int ACTC = 26;
  7 const int ACTM = 800000;
  8 const int ACQL = 800000;
  9 
 10 struct AC
 11 {
 12         int count;
 13         AC * fail;
 14         AC * ch[ ACTC ];
 15 };
 16 
 17 AC * que[ ACQL ];
 18 
 19 AC * ac_new( bool init = false ) {
 20         int i;
 21         AC * p;
 22         static AC memAC[ ACTM ];
 23         static int tot = 0;
 24         if ( init ) {
 25                 tot = 0;
 26                 return 0;
 27         }
 28         p = memAC + tot++;
 29         p->count = 0;
 30         p->fail  = 0;
 31         for ( i = 0; i < ACTC; ++i )
 32                 p->ch[ i ] = 0;
 33         return p;
 34 }
 35 
 36 int ac_add( AC * & root, const char * first, const char * last ) {
 37         AC ** p = &root;
 38         for ( ; ; ) {
 39                 if ( *== 0 ) *= ac_new();
 40                 if ( first == last ) return ++( (*p)->count );
 41                 p = &( (*p)->ch[ *first++ ] );
 42         }
 43 }
 44 
 45 void ac_build( AC * root ) {
 46         // root != 0
 47         int qh = 0, qt = 1, i;
 48         AC *pf, *pc, *pt;
 49         root->fail = 0;
 50         que[ 0 ] = root;
 51         while ( qh != qt ) {
 52                 pf = que[ qh ];
 53                 qh = ( qh + 1  ) % ACQL;
 54                 for ( i = 0; i < ACTC; ++i ) {
 55                         if ( pc = pf->ch[ i ] ) {
 56                                 for ( pt = pf->fail; pt && ( pt->ch[ i ] == 0 ); pt = pt->fail )
 57                                         ;
 58                                 pc->fail = pt ? pt->ch[ i ] : root;
 59                                 que[ qt ] = pc;
 60                                 qt = ( qt + 1 ) % ACQL;
 61                         }
 62                 }
 63         }
 64 }
 65 
 66 int ac_query( AC * root, const char * first, const char * last ) {
 67         // root != 0
 68         int ans = 0;
 69         AC *= root, *q;
 70         while ( first != last ) {
 71                 while ( p && ( p->ch[ *first ] == 0 ) ) {
 72                         p = p->fail;
 73                 }
 74                 if ( p ) {
 75                         q = p = p->ch[ *first++ ];
 76                         while ( q && ( q->count != -1 ) ) {
 77                                 ans += q->count;
 78                                 q->count = -1;
 79                                 q = q->fail;
 80                         }
 81                 }
 82                 else {
 83                         p = root;
 84                         ++first;
 85                 }
 86         }
 87         return ans;
 88 }
 89 
 90 char txt[ 1000009 ], pat[ 70 ];
 91 AC * root;
 92 
 93 int main() {
 94         int td, n;
 95         char * pc;
 96         scanf( "%d"&td );
 97         while ( td-- ) {
 98                 scanf( "%d%*c"&n );
 99                 ac_new( true );
100                 root = 0;
101                 while ( n-- ) {
102                         gets( pat );
103                         for ( pc = pat; *pc; ++pc )
104                                 *pc -= 'a';
105                         ac_add( root, pat, pc );
106                 }
107                 gets( txt );
108                 for ( pc = txt; *pc; ++pc )
109                         *pc -= 'a';
110                 ac_build( root );
111                 printf( "%d\n", ac_query( root, txt, pc ) );
112         }
113         return 0;
114 }
115 


posted on 2011-03-25 17:34 coreBugZJ 阅读(369) 评论(0)  编辑 收藏 引用 所属分类: ACM


只有注册用户登录后才能发表评论。
【推荐】超50万行VC++源码: 大型组态工控、电力仿真CAD与GIS源码库
网站导航: 博客园   IT新闻   BlogJava   知识库   博问   管理