Keywords Search

TimeLimit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K(Java/Other)

Problem Description

In the moderntime, 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 theimage, the system will match the keywords with description of image and showthe 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 willcontain one integer means how many cases will follow by.
Each case will contain two integers N means the number of keywords and Nkeywords follow. (N <= 10000)
Each keyword will only contains characters 'a'-'z', and the length will be notlonger than 50.
The last line is the description, and the length will be not longer than1000000.

Output

Print how manykeywords are contained in the description.

Sample Input

1

5

she

he

say

shr

her

yasherhs

Sample Output

3

标程：

#include<iostream>
#include<stdio.h>
#include<queue>
using namespace std;
struct node{
       node *fail,*s[27];
       int w;
}*head;
int t,n,i,j,sum;
char temp[51],str[1000001];
queue<node*> myqueue;
node *getfail(node *p,int k){
     if (p->s[k]!=NULL) return p->s[k];
     else
          if (p==head) return head;
          else return getfail(p->fail,k);
}
void built_trie(){
     node *root=head;
     node *tep;
     for (j=0;j<strlen(temp);j++){
         if (root->s[temp[j]-'a']==NULL){
            tep=new node;
            for (int k=0;k<26;k++)
                tep->s[k]=NULL;
            tep->w=0;
            tep->fail=head;
            root->s[temp[j]-'a']=tep;
            }

         root=root->s[temp[j]-'a'];
         if (j==strlen(temp)-1) root->w+=1;
         }
     return ;
}
void built_ac(){
     node *root;
     while (!myqueue.empty()) myqueue.pop();
     myqueue.push(head);
     while (!myqueue.empty()){//构造fail指针用BFS的方法扩展
           root=myqueue.front();
           myqueue.pop();
           for (j=0;j<26;j++)
               if (root->s[j]!=NULL){
                  myqueue.push(root->s[j]);
                  if (root==head) root->s[j]->fail=head;
                  else root->s[j]->fail=getfail(root->fail,j);
                  }
           }
     return ;
}
void find(){
     int len=strlen(str);
     node* tep;
     node *root=head;
     for (j=0;j<len;j++){
         while (root->s[str[j]-'a']==NULL && root!=head) root=root->fail;
         root=(root->s[str[j]-'a']==NULL)?head:root->s[str[j]-'a'];
         tep=root;
         while (tep!=head){//如果单词A出现过了，那么与他具有相同前缀的单词也都出现过了
               sum+=tep->w;
               tep->w=0;
               tep=tep->fail;
               }
         }
     return ;
     }
int main(){
    scanf("%d",&t);
    while (t--){
          sum=0;
          head=new node;
          for (i=0;i<26;i++)
              head->s[i]=NULL;
          head->fail=head;
          head->w=0;
          scanf("%d",&n);
          for (i=0;i<n;i++){
              scanf("%s",temp);
              built_trie();
              }
          built_ac();
          scanf("%s",str);
          find();
          printf("%d\n",sum);
          }
    return 0;
}