poj2503--Babelfish(字典树一水)

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

dog ogday
cat atcay
pig igpay
froot ootfray
loops oopslay

atcay
ittenkay
oopslay

cat
eh
loops

Huge input and output,scanf and printf are recommended.

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对应的，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] == '\0')
            break;
        sscanf(s,"%s %s", s1[i], s2[i]);
        gettree(head,s2[i],i);
        i++ ;
    }
    while(gets(s)!=NULL)
    {
        if(s[0] == '\0')
            break;
        k = f(head,s);
        if(k == -1)
            printf("eh\n");
        else
            printf("%s\n", s1[k]);
    }
    return 0;
}