POJ 2503 Babelfish(人造字典map)

                                                         Babelfish
Time Limit: 3000MS   Memory Limit: 65536K
Total Submissions: 40424   Accepted: 17216

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
froot ootfray
pig igpay
atcay
loops oopslay
ittenkay
oopslay

Sample Output

 
    
cat
eh
loops

Hint

Huge input and output,scanf and printf are recommended.

你刚刚从滑铁卢搬到一个大城市。这里的人们讲一门外语的一个难以理解的方言。幸运的是,你有一个字典帮助你了解他们。

输入包含字典条目多达100000,紧随其后的是一个空行,紧随其后的是一个消息的100000字。每一个字典条目包含一个英语单词是一条直线,紧随其后的是一个空间和一门外语单词。没有外国单词不止一次出现在字典里

用map对应键值,但因为不会对应字符串类键值,所以较麻烦

#include
#include
#include
#include
#include
using namespace std;
char h[100001][20];
int main()
{char a[20],b[20],c[20];
int k=1;
int p=0;
mapl;//定义容器
for(int i=0;;i++)
{
for(int i=0;;i++)
{
	scanf("%c",&a[i]);
	if(a[i]==' ')
	{
		a[i]='\0';
		break;
	}
	if(a[i]=='\n')
	{
		p=1;
		break;
	}
}
if(p==1)
break;
scanf("%s",b);
getchar();
   l[b]=k;
   strcpy(h[k++],a);

  
}
while(~scanf("%s",c))	
{if(l[c]==0)
printf("eh\n");
else
printf("%s\n",h[l[c]]);
}
}




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