Repository


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

Total Submission(s): 3735    Accepted Submission(s): 1366




Problem Description


When you go shopping, you can search in repository for avalible merchandises by the computers and internet. First you give the search system a name about something, then the system responds with the results. Now you are given a lot merchandise names in repository and some queries, and required to simulate the process.


 



Input


There is only one case. First there is an integer P (1<=P<=10000)representing the number of the merchanidse names in the repository. The next P lines each contain a string (it's length isn't beyond 20,and all the letters are lowercase).Then there is an integer Q(1<=Q<=100000) representing the number of the queries. The next Q lines each contains a string(the same limitation as foregoing descriptions) as the searching condition.


 



Output


For each query, you just output the number of the merchandises, whose names contain the search string as their substrings.


 



Sample Input 


20
ad
ae
af
ag
ah
ai
aj
ak
al
ads
add
ade
adf
adg
adh
adi
adj
adk
adl
aes
5
b
a
d
ad
s


 



Sample Output 


0
20
11
11
2

 



Source


​​2009 Multi-University Training Contest 4 - Host by HDU​​


 


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解析：对于一个长度为len的单词，将其子串[k，len)(0<=k<len)存入字典树，v[i]记录经过 i 节点的次数，对于每个单词，v[i]只记录一次。

          

注意这种数据：

2
sdasda
dadad
1
asdasa


代码：


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

const int maxn=1e6;
const int max_len=20;
int num,f[maxn][26];
int flag[maxn],v[maxn];
char s[max_len+5];

int main()
{
  //freopen("1.in","r",stdin);
  
  int n,i,j,k,x,len;
  scanf("%d",&n);
  for(x=1;x<=n;x++)
    {
      scanf("%s",s),len=strlen(s);
      for(k=0;k<len;k++)
        for(i=0,j=k;j<len;j++)
          {
            if(f[i][s[j]-'a']==0)f[i][s[j]-'a']=++num;
            i=f[i][s[j]-'a'];
            if(flag[i]!=x)flag[i]=x,v[i]++;
      }
  }
  scanf("%d",&n);
  while(n--)
    {
      scanf("%s",s),len=strlen(s);
      for(i=f[0][s[0]-'a'],j=1;i&&j<len;i=f[i][s[j++]-'a']);
      printf("%d\n",(j<len)?0:v[i]);
  }
  return 0;
}