poj_2503(map映射)

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本文提供了一道经典的编程题——Babelfish的解答思路及实现代码。该题要求通过给定的词汇表将一种虚构语言翻译成英语。文章详细解析了如何使用C++中的map来实现词汇映射，并提供了完整的源代码。

题目链接poj2503

Babelfish

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 38820		Accepted: 16578

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

Hint

Huge input and output,scanf and printf are recommended.

Source

Waterloo local 2001.09.22

/*题目大意：给定一系列外国单词，都对应着英语单词。然后给一个外国单词，要求输出相对应的英语单词。若不存在则输出eh 
  算法分析：可以根据map的映射关系 
*/
 
#include <iostream>
#include <map>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cstdio>
using namespace std;

int main() {
	map <string, bool> appear;		//记录单词是否出现 
	map <string , string > p;		//单词映射 
	
	char t;
	while (true) {
		char english[20];
		char t;				//临时字符 
		t = getchar();
		if (t == '\n')	break;		//如果是换行符，跳出 
		int i = 1;
		english[0] = t;
		while (true) {
			t = getchar();
			if (t == ' ') {
				english[i] = '\0';		
				break;
			}
			else	english[i++] = t;
		}
		string forign;
		cin >> forign;
		getchar();			//吃掉换行符 
		appear[forign] = true;
		p[forign] = english;
	}
	string word;
	while (cin >> word) {
		if (appear[word]) 	cout << p[word] << endl;
		else	cout << "eh" << endl;
	}
	return 0;
}