POJ 2503.Babelfish

Babelfish Time limit:3000MS Memory Limit:65536KB 64bit IO Format:%i64d &%i64 U Submit Status Practice POJ 2503

Description

You are just moved from Waterloo to a big city. The people speak an incomprehensible dialect of a foreign language. Fortunately, you had a dictionary to help you understand them.

Input

Input consists of 100,000 dictionary entries, followed by a blank line, followed by a message of up to 100,000 words . Each dictionary entry are a line containing an 中文版 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 and one word on each line. Each word in the input is a sequence of lowercase letters.

Output

Output is the message translated to 中文版, one word per line. Foreign words not in the dictionary should is translated as "eh".

Sample Input

Dog Ogdaycat atcaypig igpayfroot ootfrayloops oopslayatcayittenkayoopslay

Sample Output

Catehloops

Hint

Huge input and output,scanf and printf are recommended.

The meaning of the question is to translate the words

The key point is:

1. Read in Data
2. Mapping

Because the data exists on a line of two words and one word in two forms, we need to separate them when we read them again.

On the other hand, because the words need to be mapped, we can use mapto do a string-to-string mapping

Output with cout when final output (string format)

If further optimization time is required, the trie tree can be used

AC Code: GITHUB

1 /*2 By:ohyee3 Github:ohyee4 Homepage:http://www.oyohyee.com5 email:[email protected] ' e.com6 Blog:http://www.cnblogs.com/ohyee/7 8 かしこいかわいい? 9 エリーチカ! Ten to write out the хорошо code OH ~ One */ A  -#include <cstdio> -#include <algorithm> the#include <cstring> -#include <cmath> -#include <string> -#include <iostream> +#include <vector> -#include <list> +#include <queue> A#include <stack> at#include <map> - using namespacestd; -  - //DEBUG MODE - #defineDebug 0 -  in //Loops - #defineREP (n) for (int o=0;o<n;o++) to  + Const intMAXN =100005; - Const intMAXM = -; the  *InlineintRead_string (Chars[]) { $     CharC;Panax Notoginseng     inti =0; -     //While (!) ( ((c = GetChar ()) = = ") | | (c >= ' a ' &&c <= ' z ' ))) the     //if (c = = EOF) +     //return 0; A     if((c = GetChar ()) = =EOF) the         return 0; +      while(c = =' ') || (c >='a'&&c <='Z')) { -s[i++] =C; $c =GetChar (); $     } -S[i] =' /'; -     returni; the } - Wuyi BOOLDo () { the     Chartemp[maxm*2]; -map<string,string>dict; Wumap<string,string>:: iterator it; -     CharA[MAXM],B[MAXM]; About  $      while(read_string (temp)) { -         if(strcmp (temp,"") ==0) -              Break; -SSCANF (temp,"%s%s", A, b); ADICT[B] =A; +     } the  -      while(SCANF ("\n%s", a)! =EOF) { $cout << (Dict.count (a)? Dict[a]:"EH") <<"\ n"; the     } the          the  the     return false; - } in  the intMain () { the      while(Do ()); About     return 0; the}

POJ 2503.Babelfish