题目代号:UVA 11988

题目链接:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3139

You’re typing a long text with a broken keyboard. Well it’s not so badly broken. The only problem
with the keyboard is that sometimes the “home” key or the “end” key gets automatically pressed
(internally).
You’re not aware of this issue, since you’re focusing on the text and did not even turn on the
monitor! After you finished typing, you can see a text on the screen (if you turn on the monitor).
In Chinese, we can call it Beiju. Your task is to find the Beiju text.

Input
There are several test cases. Each test case is a single line containing at least one and at most 100,000
letters, underscores and two special characters ‘[’ and ‘]’. ‘[’ means the “Home” key is pressed
internally, and ‘]’ means the “End” key is pressed internally. The input is terminated by end-of-file
(EOF).

Output
For each case, print the Beiju text on the screen.

Sample Input
This_is_a_[Beiju]_text
[[]][][]Happy_Birthday_to_Tsinghua_University

Sample Output
BeijuThis_is_a__text
Happy_Birthday_to_Tsinghua_University

题目大意:'['出现相当于跳到开头输入,']'相当于跳到结尾,然后输出最终应该显示的模样。

解题思路:链表,但是链表太过于繁琐,不太方便,所以用数组模拟链表进行操作,详细参照算法竞赛与入门经典第二版,方法很巧妙,但是有点难理解,手动模拟一次操作然后思考一下就懂了。

代码:

# include <stdio.h>
# include <string.h>
# include <stdlib.h>
# include <iostream>
# include <fstream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <math.h>
# include <algorithm>
using namespace std;
# define pi acos(-1.0)
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define For(i,n,a) for(int i=n; i>=a; --i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define Fo(i,n,a) for(int i=n; i>a ;--i)
typedef long long LL;
typedef unsigned long long ULL; const int MAXM=;
char s[MAXM];
int Next[MAXM]; int main()
{
//freopen("in.txt", "r", stdin);
int cur,last;//cur为光标位置,last为显示屏最后一个字符
while(~scanf("%s",s+))
{
memset(Next,,sizeof(Next));
int len = strlen(s+);
Next[] = ;
cur = last = ;
for(int i = ; i <= len; i++)
{
if(s[i] == '[')
cur = ;
else if(s[i] == ']')
cur = last;
else
{
//模拟插入链表过程
Next[i] = Next[cur];//第i个字符指向光标位置
Next[cur] = i;//光标指向下一个字符
if(cur == last)//只有光标在当前最后一个字符位置或是遇到]后才执行
last = i;
cur = i;//移动光标
}
}
for(int i = Next[]; i != ; i = Next[i])
printf("%c",s[i]);
printf("\n");
memset(s,,sizeof(s));
}
return ;
}

UVA 11988 Broken Keyboard (a.k.a. Beiju Text)(链表)的更多相关文章

  1. UVa 11988 Broken Keyboard (a.k.a. Beiju Text)(链表)

    You're typing a long text with a broken keyboard. Well it's not so badly broken. The only problem wi ...

  2. UVA——11988 Broken Keyboard (a.k.a. Beiju Text)

    11988 Broken Keyboard (a.k.a. Beiju Text)You’re typing a long text with a broken keyboard. Well it’s ...

  3. 链表 UVA 11988 Broken Keyboard (a.k.a. Beiju Text)

    题目传送门 题意:训练指南P244 分析:链表模拟,维护链表的head和tail指针 #include <bits/stdc++.h> using namespace std; const ...

  4. UVa 11988 Broken Keyboard (a.k.a. Beiju Text)

    题目复制太麻烦了,甩个链接 http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=18693 直接模拟光标操作时间复杂度较高,所以用链 ...

  5. 暑假集训单切赛第二场 UVA 11988 Broken Keyboard (a.k.a. Beiju Text)(字符串处理)

    一开始不懂啊,什么Home键,什么End键,还以为相当于括号,[]里的东西先打印出来呢.后来果断百度了一下. 悲催啊... 题意:给定一个字符串,内部含有'['和']'光标转移指令,'['代表光标移向 ...

  6. UVa 11988 - Broken Keyboard (a.k.a. Beiju Text) 题解

    刘汝佳的题目,悲剧文本 -_-||| 这里使用vector<string>容器倒置记录数据,然后从后面输出就能够了. 难度就是不知道这种文档究竟哪里是開始输出,故此使用动态管理内存的容器比 ...

  7. UVA 11988 Broken Keyboard (a.k.a. Beiju Text) (链表,模拟)

    使用list来模拟就行了,如果熟悉list,那么这道题真是分分钟秒掉... list是双向循环链表,插入和删除操作非常快,缺点是不能像数组一样随机按下标读取. 一下是wiki上说明的相关函数:http ...

  8. uva - Broken Keyboard (a.k.a. Beiju Text)(链表)

    11988 - Broken Keyboard (a.k.a. Beiju Text) You’re typing a long text with a broken keyboard. Well i ...

  9. 破损的键盘(悲剧文本)(Broken Keyboard(a.k.a. Beiju Text),Uva 11988)

    破损的键盘(悲剧文本)(Broken Keyboard(a.k.a. Beiju Text),Uva 11988) 题意描述 你在输入文章的时候,键盘上的Home键和End键出了问题,会不定时的按下. ...

随机推荐

  1. 【Linux 网络编程】TCP/IP四层模型

    应用层.传输层.网络层.链路层 链路层:常用协议 ARP(将物理地址转化为IP地址) RARP(将IP地址转换为物理地址) 网络层(IP层):重要协议ICMP IP IGMP 传输层:重要的协议TCP ...

  2. Spring Security框架进阶、自定义登录

      1.Spring Security框架进阶 1.1 Spring Security简介 Spring Security是一个能够为基于Spring的企业应用系统提供声明式的安全访问控制解决方案的安 ...

  3. 2019.07.09 纪中_B

    错失AK记 2019.07.09[NOIP提高组]模拟 B 组 明明今天的题都很水,可就是没蒟蒻. 写题的时候: T0一眼高精(结果没切)T1看到2啊8啊果断转二进制观察,发现都是左移几位然后空出的位 ...

  4. LOJ 103 字串查找 题解

    题面 这道题是KMP的模板. KMP需要注意的细节有很多,所以把这篇文章发上来供参考: #include <bits/stdc++.h> using namespace std; char ...

  5. centos7搭建单机redis5.0

    目录 1. redis初步安装 2. 配置 3. 设置开机启动(centos6) 3.1 编写启动脚本 3.2 设置权限 3.3 启动测试 3.4 设置开机自启动 4. 设置开机启动(centos7) ...

  6. day 01 常量 注释 int(整型) 用户交互input 流程控制语句if

    python的编程语言分类(重点) if 3 > 2: 编译型: 将代码一次性全部编译成二进制,然后再执行. 优点:执行效率高. 缺点:开发效率低,不能跨平台. 代表语言:C 解释型: 逐行解释 ...

  7. __next__()

    def f1(n): m=n while True: m+=1 yield m a=f1(5) print(a.__next__()) 结果:6

  8. wex5 如何利用 百度地图 定位 和 天气插件

    引包: require("cordova!cordova-plugin-geolocation"); require("cordova!com.justep.cordov ...

  9. 一、模型验证CoreWebApi 管道方式(非过滤器处理)2(IApplicationBuilder扩展方法的另一种写法)

    一. 自定义中间件类的方式用一个单独类文件进行验证处理 Configure下添加配置 //app.AddAuthorize(); AddAuthorize因为参数(this IApplicationB ...

  10. squid代理简介

    squid代理 简单介绍一下正向代理和反向代理 标准代理:缓存静态页面,但是要实现这种方式必须在内部主机的浏览器内指明代理服务址和端口. 透明代理:不需要指明代理服务器的IP和端口 二)反向代理 可以 ...