POJ 2503 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.

 这题没写出来，网上有许多解法，字典树，哈希，map，

我觉得map很好写的，自己stl太水，我在网上看了两篇不错的map代码，虽然第二份超时了，但我觉得

我学到了很多，string 里的一些函数，第一份也让自己知道sscanf 的一些用法

#include <iostream>
#include <map>
#include <cstring>
#include <string>
#include <cstdio>
using namespace std;
char s[30], s1[15], s2[15];
string s3, tmp;
map <string, string> mp;
map <string, string> :: iterator it;
int main()
{
    while(gets(s)&&strlen(s))
    {
        sscanf(s, "%s %s", s1, s2);
        mp[s2] = s1;
    }
    while(cin >> s3)
    {
        it = mp.find(s3);
        if(it == mp.end())
            cout <<"eh"<< endl;
        else
            cout <<it ->second << endl;
    }
}

#include<iostream>
#include<cstdio>
#include<map>
#include<cstring>
using namespace std;
int main()
{
    string s;
    map<string,string>Q;
    while(getline(cin,s)&&s.length()!=0)
    {
        int blank=s.find(' ');//урЁЖ©у╦Я╣дн╩жц
        string x=s.substr(0,blank);
        string y=s.substr(blank+1);
        Q[y]=x;
    }
    string z;
    while(cin>>z)
    {
        if((Q[z].length())==0)
            cout<<"eh"<<endl;
        else
            cout<<Q[z]<<endl;
    }
    return 0;
}