Babelfish poj 2503 映射 sscanf

最新推荐文章于 2018-08-20 17:10:39 发布

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本程序解决了一个特定的翻译挑战，通过构建一个简易的双向字典，能够将一种外语翻译成英语。程序使用了高效的排序和查找算法来处理大规模的输入数据，包括多达100,000条的字典词条和相同数量级的消息单词。

Babelfish

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 41837		Accepted: 17734

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.

#include <iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
using namespace std;
struct node
{
char s1[22],s2[22];
}p[100005];
int cmp(node a,node b)
{
return strcmp(a.s2,b.s2)<0;
}
int main()
{
    int len=0;
     char str[55];
     while(gets(str))
     {
     if(str[0]=='\0')
     break;
     sscanf(str,"%s%s",p[len].s1,p[len].s2);
      len++;
     }
     sort(p,p+len,cmp);
     while(gets(str))
     {
        int l=0,r=len-1,mid,flag=1;
        while(l<=r)
        {
          mid=(l+r)/2;//mid=(l+r)>>1;
          if(strcmp(str,p[mid].s2)==0)
          {
            printf("%s\n",p[mid].s1);
            flag=0;
            break;
          }
          else if(strcmp(str,p[mid].s2)<0)
          r=mid-1;
          else
          l=mid+1;
        }
        if(flag)
        printf("eh\n");
     }
    return 0;
}

#include<cstring>
#include<iostream>
#include<map>
#include<cstdio>
#include<algorithm>
using namespace std;
int main()
{
  char a[55],c[55],b[55];
   map<string,string>map1;
   while(gets(a)&&a[0]!='\0')
   {
      sscanf(a,"%s%s",&b,&c);
      map1[c]=b;
   }
   while(gets(a)&&a[0]!='\0')
   {
       if(map1.find(a)==map1.end())
         cout<<"eh\n";
         else
         cout<<map1[a]<<endl;
   }

}