Time Limit: 10 Seconds Memory Limit: 32768 KB
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
赤裸裸的字典樹，最近寫字典樹感覺不錯，有寫了一遍，把這題搞定。
注意輸入！原來有一個輸入神器：sscanf（）
gets(str);
sscanf(str, "%s %s", ch, sh);
把字符串分成兩部分，遇到空格就分開。如果有2個空格分成三段，則上述只能賦值前兩段。 是一個用空格分開字符串賦值的神器啊！！！
鏈接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1109
題目：
#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;
}