hdu2222 Keywords Search ac自动机

地址：http://acm.split.hdu.edu.cn/showproblem.php?pid=2222

题目：

Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 56558    Accepted Submission(s): 18493

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

 

思路：ac自动机的模板题啊，我的ac自动机过的第一个题

　　不过hdu的测评机好感人，第一交代码500msAC了，，后来由于某些原因再叫同一份代码时居然TLE了！！！再交时间从600-900多的都有。。。真是醉了

 #include <queue>
 #include <cstring>
 #include <cstdio>
 using namespace std;

 struct AC_auto
 {
     const static int LetterSize = ;
     const static int TrieSize =  * ( 1e5 + );

     int tot,root,fail[TrieSize],end[TrieSize],next[TrieSize][LetterSize];

     int newnode(void)
     {
         memset(next[tot],-,sizeof(next[tot]));
         end[tot] = ;
         return tot++;
     }

     void init(void)
     {
         tot = ;
         root = newnode();
     }

     int getidx(char x)
     {
         return x - 'a';
     }

     void insert(char *ss)
     {
         int len = strlen(ss);
         int now = root;
         for(int i = ; i < len; i++)
         {
             int idx = getidx(ss[i]);
             if(next[now][idx] == -)
                 next[now][idx] = newnode();
             now = next[now][idx];
         }
         end[now]++;
     }

     void build(void)
     {
         queue<int>Q;
         fail[root] = root;
         for(int i = ; i < LetterSize; i++)
             if(next[root][i] == -)
                 next[root][i] = root;
             else
                 fail[next[root][i]] = root,Q.push(next[root][i]);
         while(Q.size())
         {
             int now = Q.front();Q.pop();
             for(int i = ; i < LetterSize; i++)
             if(next[now][i] == -)   next[now][i] = next[fail[now]][i];
             else
                 fail[next[now][i]] = next[fail[now]][i],Q.push(next[now][i]);
         }
     }

     int match(char *ss)
     {
         int len,now,res;
         len = strlen(ss),now = root,res = ;
         for(int i = ; i < len; i++)
         {
             int idx = getidx(ss[i]);
             int tmp = now = next[now][idx];
             while(tmp)
             {
                 res += end[tmp];
                 end[tmp] = ;//
                 tmp = fail[tmp];
             }
         }
         return res;
     }
     void debug(void)
     {
         for(int i = ;i < tot; i++)
         {
             printf("id = %3d,fail = %3d,end = %3d,chi = [",i,fail[i],end[i]);
             for(int j = ;j < LetterSize;j++)
                 printf("%3d",next[i][j]);
             printf("]\n");
         }
     }
 };

 AC_auto ac;
 char sa[];
 int main(void)
 {
     int t,n;
     scanf("%d",&t);
     while(t--)
     {
         ac.init();
         scanf("%d",&n);
         while(n--)
             scanf("%s",sa),ac.insert(sa);
         ac.build();
         scanf("%s",sa);
         printf("%d\n",ac.match(sa));
     }

     return ;
 }