POJ 2503 Babelfish (사전 나무)

2458 단어 데이터 구조poj
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

순수 사전 나무 문제.
배 운 지식:
① sscanf 함수, scanf 함수 의 형제, 문자열 을 지정 한 형식 으로 지정 한 변수 에 읽 습 니 다.
② 노드 노드 를 비 워 두 면 NULL 대신 0 을 사용 하면 더욱 간결 하고 안전 합 니 다.
/*
Trie(   )
*/
#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];	//   ch[]      0,     NULL     
}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
"); return ; } else u=u->ch[s[i]-'a']; } printf("%s
",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); // sscanf add(s2,s1); } while(scanf("%s",s1)!=EOF){ ask(s1); } return 0; }

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