hdu2222 ac自动机入门

Keywords Search

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 27138 Accepted Submission(s): 8871

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，模板题：

  1 #include <iostream>
  2 #include <stdio.h>
  3 #include <stdlib.h>
  4 #include <algorithm>
  5 #include <math.h>
  6 #include <queue>
  7 #include <string.h>
  8 using namespace std;
  9 typedef struct abc
 10 {
 11      abc *next[26];
 12     int end;
 13     abc *fail;
 14 }abcd;
 15 int ans;
 16 abcd *inti()
 17 {
 18     abcd *t;
 19     t=(abcd *)malloc(sizeof(abcd));
 20     t->end=0;
 21     t->fail=NULL;
 22     for(int i=0;i<26;i++)
 23     t->next[i]=NULL;
 24     return t;
 25 }
 26 void insert(abcd *t,char z[])
 27 {
 28     if(*z=='\0')
 29     {
 30         t->end++;
 31         return ;
 32     }
 33     if(t->next[*z-'a']==NULL)
 34     t->next[*z-'a']=inti();
 35     insert(t->next[*z-'a'],z+1);
 36 }
 37 void ac(abcd *t)
 38 {
 39     queue< abcd* >a;
 40     abcd *r,*f;
 41     while(!a.empty())a.pop();
 42     for(int i=0;i<26;i++)
 43     {
 44         if(t->next[i]!=NULL)
 45         a.push(t->next[i]),t->next[i]->fail=t;
 46         else
 47         t->next[i]=t;
 48     }
 49     while(!a.empty())
 50     {
 51         r=a.front();
 52         a.pop();
 53         for(int i=0;i<26;i++)
 54         {
 55             if(r->next[i]!=NULL)
 56             {
 57                 a.push(r->next[i]);
 58                 f=r->fail;
 59                 while(f->next[i] == NULL) f = f->fail;
 60                 r->next[i]->fail=f->next[i];
 61             }
 62         }
 63     }
 64 }
 65 void query(abcd *t,char x[])
 66 {
 67     abcd *p = t, *f;
 68     int i;
 69     while(*x)
 70     {
 71         i = *x - 'a';
 72         while(p->next[i] == NULL) p = p->fail;
 73         p = p->next[i];
 74         f = p;
 75         while(f != t && f->end !=-1)
 76         {
 77             ans+=f->end;
 78             f->end =-1;
 79             f = f->fail;
 80         }
 81         x++;
 82     }
 83 }
 84 void del(abcd *t)
 85 {
 86     for(int i=0;i<26;i++)
 87     {
 88         if(!t->next[i])
 89         del(t->next[i]);
 90     }
 91     free(t);
 92 }
 93 int main()
 94 {
 95     int t,n,i;
 96     cin>>t;
 97     char z[60];
 98     char x[1000009];
 99     while(t--)
100     {
101         abcd *t;
102         t=inti();
103         cin>>n;
104         for(i=0;i<n;i++)
105         {
106             cin>>z;
107             insert(t,z);
108         }
109         cin>>x;
110         ac(t);
111         ans=0;
112         query(t,x);
113         cout<<ans<<endl;
114         del(t);
115     }
116 }

View Code

posted on 2013-08-23 18:54 ERKE 阅读(284) 评论(0) 编辑收藏举报