《ACM书中题目》 B

• 原题

  Description
  We all know that FatMouse doesn’t speak English. But now he has to be prepared since our nation will join WTO soon. Thanks to Turing we have computers to help him.

  Input Specification

  Input consists of up to 100,005 dictionary entries, followed by a blank line, followed by a message of up to 100,005 words. Each dictionary entry is a line containing an English word, followed by a space and a FatMouse word. No FatMouse word appears more than once in the dictionary. The message is a sequence of words in the language of FatMouse, one word on each line. Each word in the input is a sequence of at most 10 lowercase letters.

  Output Specification

  Output is the message translated to English, one word per line. FatMouse 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

• 理解&思路
  词典类程序，给定单词与释义并查询，但要注意词条量多达100005，直接查找一定会超时。
  刚开始的想法：

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int i,n,m,z=0;
    struct dic
    {
        string x;
        string y;
    }dic[100005];
    dic[0].x="123";
    for(i=1;;i++)
    {
        cin>>dic[i].y;
        for(n=i;n>0;n--)
        {
            if(dic[i].y==dic[n].x)
            {
                cout<<dic[n].y;
                z=1;
                break;
            }
        }
        if(z==1) break;
        cin>>dic[i].x;
    }
    m=i-1;
    for(;;i++)
    {
        z=0;
        cin>>dic[i].x;
        for(n=m;n>0;n--)
        {
            if(dic[i].x==dic[n].x)
            {
                cout<<dic[n].y;
                z=1;
            }
        }
        if(z==0) cout<<"eh";
    }
}

很明显会超时，对空行的判断部分也有问题。后来看到map函数，尝试模仿使用。

• AC代码

#include<bits/stdc++.h>
using namespace std;
int main()
{
    string s;
    char ss[100],s1[100],s2[100];
    int x,y;
    map<string,string>::iterator p;
    map<string,string>m;
    while(gets(ss))
    {
        s=ss;
        if(s=="")break;
        else
        {
            sscanf(s.c_str(),"%s %s",s1,s2);
            m[s2]=s1;
        }
    }
    while(gets(ss))
    {
        s=ss;
        p=m.find(s);
        if(p!=m.end())
            cout<<m[s]<<endl;
        else
            cout<<"eh"<<endl;
    }
}

• 总结
  看解析做出来的一道题。。。之前对map函数的用法及优点不了解。