POJ Babelfish 2503(map,字典树)_pojbabelfish-优快云博客

本文介绍了一道名为Babelfish的编程题，任务是使用字典将一种外语翻译成英语。提供了两种解决方案：使用map和字典树。map方案简洁明了，而字典树则展示了更底层的数据结构实现。

Babelfish

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 40265		Accepted: 17159

Description

You have just moved from Waterloo to a big city. The people here speak an incomprehensible (费解的) dialect (方言) 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 (小写字母) 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

Hint

Huge input and output,scanf and printf are recommended.

Source

Waterloo local 2001.09.22

题意：给出单词对应关系，然后询问，输出相对应的单词

可以用map水过去

也可以用字典树

map


#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <map>
using namespace std;

int main()
{
    char a[50],s[20],str[20];
    map<string,string>k;
    while(gets(a)&&a[0]!='\0')
    {
        sscanf(a,"%s%s",str,s);
        k[s]=str;
    }
    while(gets(a) && a[0]!='\0')
    {
        sscanf(a,"%s",s);
       if(k.find(s)==k.end())
       {
           printf("eh\n");
       }
       else
       {
           cout<<k[s]<<endl;
       }
    }
    return 0;
}

字典树

#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
struct node
{
    int flag;
    int next[26];
}q[510000];
char s[100100][11];
int top=0;
int newnode()
{
    q[top].flag=0;
    memset(q[top].next,-1,sizeof(q[top].next));
    return top++;
}
void Inerst(int p,char s[],int l)
{
    //printf("%d\n",l);
    int i=0;
    for(;s[i]!='\0';i++)
    {
        int t=s[i]-'a';
        if(q[p].next[t]==-1)
        {
            q[p].next[t]=newnode();
        }
        p=q[p].next[t];
    }
    q[p].flag=l;
}
int query(int p,char s[])
{
    int i=0;
    for(;s[i]!='\0';i++)
    {
        int t=s[i]-'a';
        if(q[p].next[t]==-1)
        {
            return -1;
        }
        p=q[p].next[t];
    }
    return q[p].flag;
}

int main()
{
    char str[15];
    char a[50];
    int l=1;
    int k=newnode();
    //printf("%d\n",k);
    while(gets(a)&&a[0]!='\0')
    {
        sscanf(a,"%s%s",s[l],str);
        Inerst(k,str,l);
        l++;
    }
    while(gets(a)&&a[0]!='\0')
    {
        sscanf(a,"%s",str);
        int g=query(k,str);
        //printf("%d\n",g);
        if(g==-1)
        {
            printf("eh\n");
        }
        else
        {
            printf("%s\n",s[g]);
        }
    }

    return 0;
}