What Are You Talking About

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 102400/204800 K (Java/Others)
Total Submission(s): 11730    Accepted Submission(s): 3763

Problem Description
Ignatius is so lucky that he met a Martian yesterday. But he didn't know the language the Martians use. The Martian gives him a history book of Mars and a dictionary when it leaves. Now Ignatius want to translate the history book into English. Can you help him?
 
Input
The problem has only one test case, the test case consists of two parts, the dictionary part and the book part. The dictionary part starts with a single line contains a string "START", this string should be ignored, then some lines follow, each line contains two strings, the first one is a word in English, the second one is the corresponding word in Martian's language. A line with a single string "END" indicates the end of the directory part, and this string should be ignored. The book part starts with a single line contains a string "START", this string should be ignored, then an article written in Martian's language. You should translate the article into English with the dictionary. If you find the word in the dictionary you should translate it and write the new word into your translation, if you can't find the word in the dictionary you do not have to translate it, and just copy the old word to your translation. Space(' '), tab('\t'), enter('\n') and all the punctuation should not be translated. A line with a single string "END" indicates the end of the book part, and that's also the end of the input. All the words are in the lowercase, and each word will contain at most 10 characters, and each line will contain at most 3000 characters.
 
Output
In this problem, you have to output the translation of the history book.
 
Sample Input
 
START
from fiwo
hello difh
mars riwosf
earth fnnvk
like fiiwj
END
START
difh, i'm fiwo riwosf.
i fiiwj fnnvk!
END
 
Sample Output
 
hello, i'm from mars.
i like earth!
 
 

Huge input, scanf is recommended.

 
 
 
 /*
模板题 */
#include<iostream>
#include<stdio.h>
#include<cstring>
#include<cstdlib>
using namespace std; struct Tril
{
char a[];
struct Tril *next[];
};
Tril *root;
void mem(Tril *p)
{
int i;
for(i=;i<;i++) p->next[i]=NULL;
p->a[]='\0';
} void insert_Tril(char *s1,char *s2)
{
int i,k,ans;
struct Tril *p=root,*q;
k=strlen(s2);
for(i=;i<k;i++)
{
ans=s2[i]-'a';
if( p->next[ans]==NULL )
{
q=(struct Tril*)malloc(sizeof(struct Tril));
mem(q);
p->next[ans]=q;
p=q;
}
else
{
p=p->next[ans];
}
}
strcpy(p->a,s1);
}
void found_Tril(char *s1)
{
int i,k,ans;
struct Tril *p=root;
bool flag=false;
k=strlen(s1);
for(i=;i<k;i++)
{
ans=s1[i]-'a';
if(p->next[ans]==NULL)
{
flag=true;
break;
}
else p=p->next[ans];
}
if(flag==true || strlen(p->a)==)
printf("%s",s1);
else printf("%s",p->a);
}
int main()
{
int i,k;
char s1[],s2[];
root=(struct Tril*)malloc(sizeof(struct Tril));
mem(root);
scanf("%s",s1);
while(scanf("%s",s1))
{
if( strcmp(s1,"END")==)break;
scanf("%s",s2);
insert_Tril(s1,s2);
}
scanf("%s",s1);
getchar();
while(gets(s1))
{
if( strcmp(s1,"END")==)break;
i=;
while(s1[i]!='\0')
{
if( s1[i]>'z'||s1[i]<'a')
{
printf("%c",s1[i]);
i++;
}
else
{
k=;
while(s1[i]<='z'&&s1[i]>='a' &&s1[i]!='\0')
{
s2[k]=s1[i];
i++;
k++;
}
s2[k]='\0';
found_Tril(s2);
}
}
printf("\n");
}
return ;
}
 

hdu 1075 What Are You Talking About 字典树模板的更多相关文章

  1. 字典树模板题(统计难题 HDU - 1251)

    https://vjudge.net/problem/HDU-1251 标准的字典树模板题: 也注意一下输入方法: #include<iostream> #include<cstdi ...

  2. HDU - 1251 字典树模板题

    Ignatius最近遇到一个难题,老师交给他很多单词(只有小写字母组成,不会有重复的单词出现),现在老师要他统计出以某个字符串为前缀的单词数量(单词本身也是自己的前缀).  Input输入数据的第一部 ...

  3. 字典树模板 HDU - 1251

    题意: 给一些单词,换行键后,查找以后输入的单词作为前缀的话们在之前出现过几次. 思路: 字典树模板----像查字典的顺序一样 #include<string> #include<s ...

  4. hdu1521(字典树模板)

    题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1251 题意: 中文题诶~ 思路: 字典树模板 代码1: 动态内存, 比较好理解一点, 不过速度略慢, ...

  5. CH 1601 - 前缀统计 - [字典树模板题]

    题目链接:传送门 描述给定 $N$ 个字符串 $S_1,S_2,\cdots,S_N$,接下来进行 $M$ 次询问,每次询问给定一个字符串 $T$,求 $S_1 \sim S_N$ 中有多少个字符串是 ...

  6. HDU 4825 Xor Sum(经典01字典树+贪心)

    Xor Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others) Total ...

  7. HDU 2072 - 单词数 - [(有点小坑的)字典树模板题]

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2072 Problem Descriptionlily的好朋友xiaoou333最近很空,他想了一件没有 ...

  8. HDU 1251 统计难题(字典树模板题)

    http://acm.hdu.edu.cn/showproblem.php?pid=1251 题意:给出一些单词,然后有多次询问,每次输出以该单词为前缀的单词的数量. 思路: 字典树入门题. #inc ...

  9. HDU:1251-统计难题(字典树模板,动态建树,静态建树)

    传送门:http://acm.hdu.edu.cn/showproblem.php?pid=1251 统计难题 Time Limit: 4000/2000 MS (Java/Others) Memor ...

随机推荐

  1. java项目迁移

    电脑重装系统以后或者从不同MyEclipse版本迁移项目时候会出现: Project facet Java 1.5 is not supported by target runtime Apache ...

  2. solr安装教程

    Solr Solr is the popular, blazing-fast, open source enterprise search platform built on Apache Lucen ...

  3. D01——C语言基础学PYTHON

    C语言基础学习PYTHON——基础学习D01 20180705内容纲要: 1 PYTHON介绍 2 PYTHON变量定义规则 3  PYTHON文件结构 4 PYTHON语句及语法 5 字符编码 6 ...

  4. Hibernate中连接数据库的配置

    Hibernate连接数据库的配置 实体类的映射文件 <!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mappin ...

  5. 论文笔记:CNN经典结构2(WideResNet,FractalNet,DenseNet,ResNeXt,DPN,SENet)

    前言 在论文笔记:CNN经典结构1中主要讲了2012-2015年的一些经典CNN结构.本文主要讲解2016-2017年的一些经典CNN结构. CIFAR和SVHN上,DenseNet-BC优于ResN ...

  6. 前端知识总结--js原型链

    js的原型链听着比较深奥,看着容易晕,梳理一下还是比较容易懂的 (先简单写下,后续有时间再整理) 简而言之 原型链:就是js的对象与对象之间,通过原型组成建立的层层关系,构成了整个链条,称之为原型链  ...

  7. 再也不怕aop的原理了

    1 aop是什么 java的核心思想是面向对象,aop是面向切面编程.是对面向对象的一个补充,简单通俗的理解下aop,假设我们上楼梯,我的目标是三楼,我直接朝我的三楼直接过去,但是我可以在二楼的时候去 ...

  8. 深入浅出理解基于 Kafka 和 ZooKeeper 的分布式消息队列

    消息队列中间件是分布式系统中重要的组件,主要解决应用耦合,异步消息,流量削锋等问题.实现高性能,高可用,可伸缩和最终一致性架构,是大型分布式系统不可缺少的中间件. 本场 Chat 主要内容: Kafk ...

  9. 图像的上采样(upsampling)与下采样(subsampled)

    缩小图像(或称为下采样(subsampled)或降采样(downsampled))的主要目的有两个:1.使得图像符合显示区域的大小:2.生成对应图像的缩略图. 放大图像(或称为上采样(upsampli ...

  10. c# 抽象类 抽象方法

    抽象类与非抽象类的主要区别: ·抽象类不能直接被实例化 ·抽象类中可以包含抽象成员,但非抽象类中不可以 ·抽象类不能被密封 声明抽象方法时需注意:·抽象方法必须声明在抽象类中 ·声明抽象方法时,不能使 ...