POJ2503Babelfish(MAP+字典树)

最新推荐文章于 2021-03-20 00:18:40 发布

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本文探讨了两种解决特定翻译问题的方法：使用标准的Map数据结构和利用字典树（Trie）。通过对比两种实现方式的性能，文章详细介绍了如何构建字典树以高效地完成翻译任务，并附带提供了完整的源代码。

Babelfish

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 37423		Accepted: 15967

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

题意：输入的每一单词跟每一个单词对应，查找第二个单词对应的第一个单词

直接map<string,string>建立关系，后来输出的时候scanf是不可以的，因为map里面的是string

然后看到有说用字典树写的，然后又写了个字典树，没调试，直接交，水过

//map 2000ms，原因可能是后面用的cout
#include <iostream>
#include <cstring>
#include <string.h>
#include <cstdio>
#include <map>
#include <ctype.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
int main()
{
    map<string,string> g;
    char str[30],s1[15],s2[15];
    while(gets(str) && strcmp(str,"") != 0)
    {
        sscanf(str,"%s %s",s1,s2);
        g[s2] = s1;
    }
    while(gets(s2))
    {
        if(g.count(s2) == 0)
            printf("eh\n");
        else
            cout << g[s2]<<endl;
    }
    return 0;
}

字典树 735ms

#include <iostream>
#include <cstring>
#include <string.h>
#include <cstdio>
#include <map>
#include <ctype.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
struct node
{
    int flag;
    node *next[26];
    char s[15];
};
node *root;
void build(char *s2,node *root,char *s1)
{
    node *p,*q;
    p = root;
    int len = strlen(s2);
    for(int i = 0; i < len; i++)
    {
        int id = s2[i] - 'a';
        if(p -> next[id] == NULL)
        {
            q = (node *) malloc(sizeof(node));
            for(int i = 0; i < 26; i++)
                q -> next[i] = NULL;
            q -> flag = 0;
            p -> next [id] = q;
            p = p -> next[id];
        }
        else
            p = p -> next[id];
    }
    p -> flag = 1;
    strcpy(p -> s,s1);
}
void Find(node *root,char *s2)
{
    node *p;
    p = root;
    int len = strlen(s2);
    for(int i = 0; i < len; i++)
    {
        int id = s2[i] - 'a';
        if(p -> next[id] == NULL)
        {
            printf("eh\n");
            return ;
        }
        else
            p = p -> next[id];
    }
    if(p -> flag)
        printf("%s\n",p -> s);
    else
        printf("eh\n");
}
int main()
{
    char str[30],s1[15],s2[15];
    root = (node *)malloc(sizeof(node));
    for(int i = 0; i < 26; i++)
        root->next[i] = NULL;
    while(gets(str) && strcmp(str,"") != 0)
    {
        sscanf(str,"%s %s",s1,s2);
        build(s2,root,s1);
    }
    while(gets(s2))
    {
        Find(root,s2);
    }
    return 0;
}