ZOJ 1109 Language of FatMouse(赤裸裸的字典树)

Language of FatMouse

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Time Limit: 10 Seconds      Memory Limit: 32768 KB

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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

Output for Sample Input

cat
eh
loops

 

          赤裸裸的字典树，最近写字典树感觉不错，有写了一遍，把这题搞定。

       注意输入！原来有一个输入神器：sscanf（）

       gets(str);

       sscanf(str, "%s %s", ch, sh);

       把字符串分成两部分，遇到空格就分开。如果有2个空格分成三段，则上述只能赋值前两段。 是一个用空格分开字符串赋值的神器啊！！！

 

链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1109

题目：

#include <iostream>
#include <stdio.h>
#include <cstring>
#include <string>
using namespace std;

struct node
{
    char ch[12];
    node *next[26];
};

node *root, memory[200005];
int cnt;

node *create()
{
    node *p = &memory[cnt++];
    int i;
    for(i = 0; i < 26; i++)
        p->next[i] = NULL;
    return p;
};

void insert(char *s, char *c)
{
    node *p = root;
    int i, k;
    for(i = 0; s[i]; i++)
    {
        k = s[i] - 'a';
        if(p->next[k] == NULL)
        {
            p->next[k] = create();
        }
        p = p->next[k];
    }
    strcpy(p->ch, c);
}

void search(char *s)
{
    node *p = root;
    int i, k;
    for(i = 0; s[i]; i++)
    {
        k = s[i] - 'a';
        if(p->next[k] == NULL)
        {
            printf("eh\n");
            return;
        }
        p = p->next[k];
    }
    printf("%s\n", p->ch);
}

int main()
{
    char ch[15], sh[15], str[35];
    root = create();
    while(gets(str))
    {
        if(str[0] == 0) break;
        sscanf(str, "%s %s", ch, sh);
        insert(sh, ch);
    }
    while(gets(ch))
    {
        search(ch);
    }

    return 0;
}