Babelfish
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 35009 | Accepted: 14979 |
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
纯字典树水题。
学到的知识：
① sscanf 函数，scanf函数的兄弟，把字符串按指定格式读入到指定变量。
② Node 节点置空，可以不用NULL，而是用0更简洁安全。
/*
Trie(字典树)
*/
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
typedef __int64 ll;
typedef struct Node{
char word[15];
struct Node *ch[27]; //没想到ch[]的初值可以为0，一开始我用NULL异常退出了
}node;
node *T;
int index;
void add(char s[],char wd[]){
node *u;
u=T;
int len=strlen(s),i;
for(i=0;i<len;i++){
if(u->ch[s[i]-'a']==0){
node *t=(node*)malloc(sizeof(node));
for(int j=0;j<27;j++)
t->ch[j]=0;
u->ch[s[i]-'a']=t;
}
u=u->ch[s[i]-'a'];
}
strcpy(u->word,wd);
}
void ask(char s[]){
node* u=T;
int i,len=strlen(s);
for(i=0;i<len;i++){
if(u==0){
printf("eh\n");
return ;
}
else
u=u->ch[s[i]-'a'];
}
printf("%s\n",u->word);
}
int main()
{
int i,k,j;
char s1[20],s2[20],str[50];
T=(node*)malloc(sizeof(node));
for(i=0;i<27;i++)
T->ch[i]=0;
while(gets(str)){
if(str[0]=='\0') break;
sscanf(str,"%s %s",s1,s2); //需要借助sscanf来处理
add(s2,s1);
}
while(scanf("%s",s1)!=EOF){
ask(s1);
}
return 0;
}