hdu 2846 Repository

http://acm.hdu.edu.cn/showproblem.php?pid=2846

Repository

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2720    Accepted Submission(s): 1060

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

 

分析：

 

典型的字典树问题。

 

AC代码：

 

 #include<stdio.h>
 #include<iostream>
 #include<string.h>

 using namespace std;

 struct node
 {
     int count;
     int value;
     node *next[];
     node()
     {
         for(int i=;i<;i++)
             next[i]=;
         count=;
         value=;
     }
 };

 void insert(node *&root,char *s,int num)
 {
     int i=,t=;
     node *p=root;
     if(!p)
     {
         p=new node();
         root=p;
     }
     while(s[i])
     {
         t=s[i]-'a';
         if(!p->next[t])
             p->next[t]=new node();
         p=p->next[t];
         if(p->value!=num)
         {
             p->count++;
             p->value=num;
         }
     //    printf("%d\n",p->value);
         i++;
     }
 }

 int find(node *root,char *s)
 {
     int i=,t=,ans;
     node *p=root;
     if(!p)
         return ;
     while(s[i])
     {
         t=s[i]-'a';
         if(!p->next[t])
             return ;
         p=p->next[t];
         ans=p->count;
         i++;
     }
     return ans;
 }

 int main()
 {
     int T;
     scanf("%d",&T);
     char str[];
     node *root=NULL;
     while(T--)
     {
         scanf("%s",str);
         for(int i=;i<strlen(str);i++)
             insert(root,str+i,T+);
     }
     int Q;
     scanf("%d",&Q);
     while(Q--)
     {
         scanf("%s",str);
         printf("%d\n",find(root,str));
     }
     return ;
 }