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  • poj2503--Babelfish(特里一水)

    Babelfish
    Time Limit: 3000MS   Memory Limit: 65536K
    Total Submissions: 32988   Accepted: 14189

    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.

    Source

    Waterloo local 2001.09.22
    输入一个字典。前面一个单词和后面的单词映射。问给出的单词有没有相应的,有输出,没有,输出eh
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    using namespace std;
    struct node{
        int flag ;
        node *next[27] ;
    } *head;
    node *getnode()
    {
        node *p = new node ;
        int i ;
        for(i = 0 ; i < 27 ; i++)
            p->next[i] = NULL ;
        p->flag = -1 ;
        return p ;
    }
    void gettree(node *p,char *s,int m)
    {
        int i , k , l = strlen(s);
        for(i = 0 ; i < l ; i++)
        {
            k = s[i] - 'a' ;
            if( p->next[k] == NULL )
                p->next[k] = getnode();
            p = p->next[k] ;
        }
        p->flag = m ;
    }
    int f(node *p,char *s)
    {
        int i , k , l = strlen(s) ;
        for(i = 0 ; i < l ; i++)
        {
            k = s[i] - 'a' ;
            if( p->next[k] == NULL )
                return -1 ;
            p = p->next[k] ;
        }
        return p->flag;
    }
    char s1[110000][12] , s2[110000][12] , s[30] ;
    int main()
    {
        int i = 0 , j , l , k ;
        head = getnode();
        while(1)
        {
            gets(s);
            if(s[0] == '')
                break;
            sscanf(s,"%s %s", s1[i], s2[i]);
            gettree(head,s2[i],i);
            i++ ;
        }
        while(gets(s)!=NULL)
        {
            if(s[0] == '')
                break;
            k = f(head,s);
            if(k == -1)
                printf("eh
    ");
            else
                printf("%s
    ", s1[k]);
        }
        return 0;
    }
    

    版权声明:转载请注明出处:http://blog.csdn.net/winddreams

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  • 原文地址:https://www.cnblogs.com/hrhguanli/p/4846000.html
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