POJ-2503-Babelfish

最新推荐文章于 2021-08-21 18:22:26 发布

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本文介绍了一个使用字典树（Trie）解决语言翻译问题的编程挑战。通过构建字典树来高效地进行词汇查找，实现从一种语言到另一种语言的翻译功能。文章提供了完整的AC代码示例，并对比了字典树与map的不同之处。

Babelfish
Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 47818 Accepted: 20030
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
解题思路： 这道题用map也可以做,但我最近学的字典树，所以用的字典树做的，map的做法比较简单，但空间消耗也比较大，这道题是我写的第一个用指针开些字典树的题，之前我用的都是数组模拟，但是在后来提交的时候总是RE，不明白为什么，后来看同学用的指针，，大一的时候指针没学好，就来学学，指针和数组模拟的思路是一样，但是指针空间消耗比较小，具体关于字典树的模板我就不讲了，如果不了解，可以看看网上大佬写的关于字典树的解释，很清晰。
AC代码

#include<iostream>
#include<Stdio.h>
#include<stdlib.h>
#include<string.h>
using namespace std;
#define maxn 27
typedef struct TrieNode
{
    char s[10];
    TrieNode *son[maxn];
}Trie;
void insertWord(Trie *root,char str1[],char str2[])
{
    Trie *p=root;
    int i=0;
    char var=str2[i];
    while(var!='\0')
    {

        if(p->son[var-'a']==NULL)
        {
            Trie *node=(Trie *)malloc(sizeof(Trie));
            for(int j=0;j<maxn;j++)
            {
                node->son[j]=NULL;
            }
            p->son[var-'a']=node;
            p=p->son[var-'a'];
        }
        else
        {
            p=p->son[var-'a'];
        }
        var=str2[++i];
    }
    strcpy(p->s,str1);
}
void findWord(Trie *root,char word[])
{
    int i=0;
    char ch=word[i];
    Trie *p=root;
    while(p!=NULL&&ch!='\0')
    {
        p=p->son[ch-'a'];
        ch=word[++i];
    }
    if(p==NULL)
        printf("eh\n");
    else
        printf("%s\n",p->s);
}
void freeTrie(Trie *root)
{
    for(int i=0;i<maxn;i++)
    {
        if(root->son[i]!=NULL)
            freeTrie(root->son[i]);
    }
    free(root);
}
int main()
{
    char ch[30],s1[11],s2[11];
    Trie *root=(Trie *)malloc(sizeof(Trie));
    for(int i=0;i<maxn;i++)
    {
        root->son[i]=NULL;
    }
    while(gets(ch))
    {
        if(strlen(ch)==0)
        {

            break;
        }
        sscanf(ch,"%s %s",s1,s2);
        insertWord(root,s1,s2);
    }
    while(gets(ch))
    {
        findWord(root,ch);
    }
    freeTrie(root);
    return 0;
}