杭電2846

Repository

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2109    Accepted Submission(s): 779

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

 

這個題目我開始想法就是先把樹建起來，然後再在樹中查找，但發現查找的那個函數會很複雜。。

借鑑其他人的方法是：把每個字符串的每一個子串在樹中建立起來，那麼查找的時候就會相對方便一些

#include <cstdio>
#include <cstring>
#include <cstdlib>
struct node
{
    node *x[26];
    int count;
    int num;
};
node *tree;
char stu[30];
int p,q;
int flag;
void create(node *&t)
{
    int i;
    t=(node *)malloc(sizeof(node));
    t->count=0;
    t->num=-1;
    for(i=0;i<26;++i)
        t->x[i]=NULL;
}
void insertTree(char *ch,int n)
{
  //  printf("ch %s  n %d\n",ch,n);
   node *f=tree;
   int i,j;
   int len=strlen(ch);
   for(i=0;i<len;++i)
   {
       int p=ch[i]-'a';
       if(f->x[p]==NULL)
       {
           f->x[p]=(node *)malloc(sizeof(node));
           f=f->x[p];
           f->count=0;
           f->num=-1;
           for(j=0;j<26;++j)
                 f->x[j]=NULL;
            if(i==len-1)
            {
                f->count=1;
                f->num=n;
            }
       }
       else
       {
           f=f->x[p];
           if((i==len-1) && (f->num!=n))
           {
               f->num=n;
               f->count++;
           }
       }
   }
}
int chaxun()
{
    node *f=tree;
   int i,len=strlen(stu);
  // printf("stu %s  len %d\n",stu,len);
   for(i=0;i<len;++i)
   {
       int p=stu[i]-'a';
       if(f->x[p]==NULL)
        return 0;
       f=f->x[p];
       if(i==len-1)
        return f->count;

   }

}
void deleteTree(node *&tree)
{
   node *f=tree;
   int i;
   for(i=0;i<26;++i)
   {
       if(f->x[i]!=NULL)
       {
           deleteTree(f->x[i]);
       }
   }
   free(f);
}
int main()
{
    int i,j,k,len;
    char ch[30];
    create(tree);
    scanf("%d",&p);getchar();
 //   printf("tree  %d\np  %d\n",tree,p);
    for(i=1;i<=p;++i)
    {
        scanf("%s",stu);getchar();
        len=strlen(stu);
        int u;
        for(j=0;j<len;++j)
        {
            u=-1;
            for(k=j;k<len;++k)
            {
                ++u;
                ch[u]=stu[k];
                ch[u+1]='\0';
                insertTree(ch,i);
            }
        }
    }
    scanf("%d",&q);getchar();
 //   printf("q  %d\n",q);
    for(i=1;i<=q;++i)
    {
        scanf("%s",stu);getchar();
        flag=chaxun();
        printf("%d\n",flag);
    }
    deleteTree(tree);
    return 0;
}