Babelfish 字典树 查询O(1)

Problem Description

You have just moved from Waterloo to a big city. The people here speak an incomprehensible dialect of a foreign language. Fortunately, you have a dictionary to help you understand them.

 

Input

Input consists of up to 100,000 dictionary entries, followed by a blank line, followed by a message of up to 100,000 words. Each dictionary entry is a line containing an English word, followed by a space and a foreign language word. No foreign word appears more than once in the dictionary. The message is a sequence of words in the foreign language, one word on each line. Each word in the input is a sequence of at most 10 lowercase letters.

 

Output

Output is the message translated to English, one word per line. Foreign words not in the dictionary should be translated as "eh".

 

Sample Input

dog ogday cat atcay pig igpay froot ootfray loops oopslay atcay ittenkay oopslay

 

Sample Output

cat eh loops

***************************************************************************************************************************

字典树，无技巧

***************************************************************************************************************************

#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<cstdio>
using namespace std;
class nodea//建字典树的的类
{
  public:
    int id;
    nodea*p[27];
    nodea()
    {
        int it;
        id=-1;
        for(it=0;it<=26;it++)
          p[it]=NULL;
    }
};
nodea *root;
int ans;
int n,m;
char str1[100001][11];
char str2[100001][11];
int get_ans(char *s)//字典树的编码
{
    nodea *r=root;
    int m=strlen(s);
    for(int it=0;it<m;it++)
    {
        if(r->p[s[it]-'a']==NULL)//分支要尽可能的少
          r->p[s[it]-'a']=new nodea();
        r=r->p[s[it]-'a'];
    }
    if(r->id==-1)
    {
        r->id=ans;
        ans++;
    }
    return r->id;
}
int main()
{
    char str3[100001];
    int i,j,k,is,js;
    i=0;
    ans=0;
    root=new nodea();
    while(gets(str3))
    {
        if(strlen(str3)==0)
            break;
        j=0;
        while(str3[j]!=' ')
        {
            str1[i][j]=str3[j];
            j++;
        }
        str1[i][j]='';
        j++;
        k=0;
        while(str3[j]!='')
         {
             str2[i][k]=str3[j];
             k++;
             j++;
         }
        str2[i][j]='';
        int st=get_ans(str2[i]);
        i++;

    }
    n=ans;
   //for(j=0;j<ans;j++)
    //printf("%s%s",str1[j],str2[j]);
   while(gets(str3))
   {
       if(strlen(str3)==0)
         break;
       //for(i=0;i<strlen(str3);i++)
        //str4[i]=str3[i]-'a';
       k=get_ans(str3);
        if(k>=n)
        {
            printf("eh
");
            continue;
        }
       printf("%s
",str1[k]);//查询O(1);
   }
   return 0;
}