POJ 2503 Babelfish map做法

最新推荐文章于 2020-10-16 00:47:54 发布

原创最新推荐文章于 2020-10-16 00:47:54 发布 · 509 阅读

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本文介绍了一道名为Babelfish的编程题，任务是从一个大型字典中翻译一种虚构外语到英语。输入包括一个最大10万条目的字典和一段同样长度的待翻译文本，需要使用C++中的map来实现高效查找。

Babelfish

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 31965		Accepted: 13732

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

map的基础题。比较大的输入输出，所以重载new，delete运算符。但其实这题输入输出也不是很大，没什么必要。

代码如下：

#include <iostream>
#include<cstdio>
#include<map>
#include<string>
using namespace std;

//手动分配内存空间
char buf[50000000];
int sizet =0;

void* operator new(size_t size){
void *p = buf + sizet;
sizet += size;
return p;

}
void operator delete(void* p){
}

//
int main()
{
char e[11],f[11],str[25];
map<string,string>t;
for(;;)
{
gets(str);
//cin>>str;
if(str[0]=='\0') break;//str[0]=='\n'无效,应该是gets和cin等的运行机制问题。若要保险，再加一句：str[0]=='\n'.
sscanf(str,"%s%s",e,f);//从str读取
t[f]=e;
}
while(scanf("%s",str)!=EOF)
{
string s=t[str];
if (s.length()==0)//string类对象s的函数length（）
printf("eh\n");
else
printf("%s\n",s.c_str());//转换为cstring类型的数据，用s.c_str()函数
}
return 0;
}