poj 2503 Babelfish——字典映射

最新推荐文章于 2025-05-28 20:26:50 发布

five_east_west

最新推荐文章于 2025-05-28 20:26:50 发布

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分类专栏： ACM

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这篇博客讲述了主人公从Waterloo迁移到一个使用陌生方言的大城市，通过Babelfish字典来理解和适应新环境的故事。输入包含最多100,000条字典条目，每条由一个英文单词和对应的外语单词组成。接下来是一段外语消息，需要使用字典将其翻译成英文。输出为翻译后的消息，未找到对应翻译的单词用'eh'表示。博主分享了C++实现的解决方案，该方案能够处理大输入和输出，并提供了样例输入和输出。

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Babelfish

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.

题意： 根据给出的词典对应关系，将下面输入的词翻译成英文词汇。

题解： 字典映射，输入的处理麻烦一点，题目本身不难

c++ AC　代码

#include <map>
#include <cstdio>
#include <cstdlib>
#include <string>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1e3 + 10;

int N, K;
string arr[maxn];
int cnt = 0;

int main()
{
	char s1[15], s2[15], s[15];
	map<string, string> dict;
	while (1)
	{
		char c;

		if ((c = getchar()) == '\n')
			break;
		else
		{
			s1[0] = c;
			int i = 1;
			while (1)
			{
				c = getchar();
				if (c == ' ')
				{
					s1[i] = '\0';
					break;
				}
				else
					s1[i++] = c;
			}
		}

		cin >> s2;
		getchar();

		dict[s2] = s1;
	}

	while (cin >> s)
	{
		if (dict[s].empty())
			puts("eh");
		else
			cout << dict[s] << endl;
	}
	// system("pause");
	return 0;
}