zoukankan      html  css  js  c++  java
  • POJ

    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
    
    Hint
    Huge input and output,scanf and printf are recommended.


    代码:

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <cstdlib>
    #include <algorithm>
    
    using namespace std;
    
    char s[35],s1[15],s2[15];
    
    struct Node{
    	char word[15];
    	Node* Next[26];
    	Node(){
    		memset(Next,NULL,sizeof Next);
    	}
    };
    
    void Insert(Node* root){
    	Node* p = root;
    	int len = strlen(s2);
    	for(int i=0 ; i<len ; ++i){
    		int t = s2[i] - 'a';
    		if(p->Next[t] == NULL)p->Next[t] = new struct Node();
    		p = p->Next[t];
    	}
    	strcpy(p->word,s1);
    }
    
    void Judge(Node* root){
    	int len = strlen(s2);
    	Node* p = root;
    	for(int i=0 ; i<len ; ++i){
    		int t = s2[i]-'a';
    		if(p->Next[t])p = p->Next[t];
    		else {
    			printf("eh
    ");
    			return;
    		}
    	}
    	printf("%s
    ",p->word);
    }
    
    int main(){
    	
    	Node* root = new struct Node();
    	while(gets(s) && strlen(s)!=0){
    		sscanf(s,"%s %s",s1,s2);
    		Insert(root);
    	}
    	while(scanf("%s",s2)!=EOF){
    		Judge(root);
    	}
    	
    	return 0;
    } 

  • 相关阅读:
    C++之路进阶——codevs2439(降雨量)
    C++之路进阶——codevs2933(诗人小G)
    C++之路进阶——bzoj2879(美食节)
    C++之路进阶——bzoj1934(善意的投票)
    C++之路进阶——bzoj3876(支线剧情)
    C++之路进阶——codevs1281(Xn数列)
    八数码难题
    道路游戏
    细胞分裂
    最长链
  • 原文地址:https://www.cnblogs.com/vocaloid01/p/9514073.html
Copyright © 2011-2022 走看看