VK Cup 2012 Qualification Round 1---C. Cd and pwd commands
3 seconds
256 megabytes
standard input
standard output
Vasya is writing an operating system shell, and it should have commands for working with directories. To begin with, he decided to go with just two commands: cd (change
the current directory) and pwd (display the current directory).
Directories in Vasya's operating system form a traditional hierarchical tree structure. There is a single root directory, denoted by the slash character "/".
Every other directory has a name — a non-empty string consisting of lowercase Latin letters. Each directory (except for the root) has a parent directory — the one that contains the given directory. It is denoted as "..".
The command cd takes a single parameter, which is a path in the file system. The command changes the current directory to the directory specified by the path.
The path consists of the names of directories separated by slashes. The name of the directory can be "..", which means a step up to the parent directory. «..»
can be used in any place of the path, maybe several times. If the path begins with a slash, it is considered to be an absolute path, that is, the directory changes to the specified one, starting from the root. If the parameter begins with a directory name
(or ".."), it is considered to be a relative path, that is, the directory changes to the specified directory, starting from the current one.
The command pwd should display the absolute path to the current directory. This path must not contain "..".
Initially, the current directory is the root. All directories mentioned explicitly or passed indirectly within any command cd are considered to exist. It is guaranteed
that there is no attempt of transition to the parent directory of the root directory.
The first line of the input data contains the single integer n (1 ≤ n ≤ 50)
— the number of commands.
Then follow n lines, each contains one command. Each of these lines contains either command pwd,
or command cd, followed by a space-separated non-empty parameter.
The command parameter cd only contains lower case Latin letters, slashes and dots, two slashes cannot go consecutively, dots occur only as the name of a parent
pseudo-directory. The command parameter cd does not end with a slash, except when it is the only symbol that points to the root directory. The command parameter
has a length from 1 to 200 characters, inclusive.
Directories in the file system can have the same names.
For each command pwd you should print the full absolute path of the given directory, ending with a slash. It should start with a slash and contain the list of
slash-separated directories in the order of being nested from the root to the current folder. It should contain no dots.
7
pwd
cd /home/vasya
pwd
cd ..
pwd
cd vasya/../petya
pwd
/
/home/vasya/
/home/
/home/petya/
4
cd /a/b
pwd
cd ../a/b
pwd
/a/b/
/a/a/b/
解题思路:用字符串操作模拟树。 找绝对路径。
AC代码:
#include <iostream>
#include <cstdio>
#include <string>
using namespace std; int main(){
// freopen("in.txt", "r", stdin);
int n;
string cmd, cur, ans, tt;
while(cin>>n){
ans = "/";
for(int i=0; i<n; i++){
cin>>cmd;
if(cmd == "cd"){
cin>>cur;
cur += '/';
for(int j=0; j<cur.size(); j++){
tt += cur[j];
if(cur[j] == '/'){
if(tt == "/") ans = tt;
else if(tt == "../"){
int k;
for(k=ans.size()-1; ans[k-1]!='/'; k--) ;
ans.resize(k);
}
else ans += tt;
tt = "";
}
}
}
else{
cout<<ans<<endl;
}
}
}
return 0;
}
VK Cup 2012 Qualification Round 1---C. Cd and pwd commands的更多相关文章
- VK Cup 2012 Qualification Round 1 C. Cd and pwd commands 模拟
C. Cd and pwd commands Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/problemset ...
- DP VK Cup 2012 Qualification Round D. Palindrome pairs
题目地址:http://blog.csdn.net/shiyuankongbu/article/details/10004443 /* 题意:在i前面找回文子串,在i后面找回文子串相互配对,问有几对 ...
- VK Cup 2012 Qualification Round 2 C. String Manipulation 1.0 字符串模拟
C. String Manipulation 1.0 Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 codeforces.com/problemset/pr ...
- VK Cup 2012 Qualification Round 1 E. Phone Talks —— DP
题目链接:http://codeforces.com/contest/158/problem/E E. Phone Talks time limit per test 3 seconds memory ...
- VK Cup 2016 - Qualification Round 2 B. Making Genome in Berland
今天在codeforces上面做到一道题:http://codeforces.com/contest/638/problem/B 题目大意是:给定n个字符串,找到最短的字符串S使得n个字符串都是这个字 ...
- VK Cup 2015 - Qualification Round 1 D. Closest Equals 离线+线段树
题目链接: http://codeforces.com/problemset/problem/522/D D. Closest Equals time limit per test3 secondsm ...
- VK Cup 2016 - Qualification Round 2 D. Three-dimensional Turtle Super Computer 暴力
D. Three-dimensional Turtle Super Computer 题目连接: http://www.codeforces.com/contest/638/problem/D Des ...
- VK Cup 2016 - Qualification Round 2 C. Road Improvement dfs
C. Road Improvement 题目连接: http://www.codeforces.com/contest/638/problem/C Description In Berland the ...
- VK Cup 2016 - Qualification Round 2 B. Making Genome in Berland 水题
B. Making Genome in Berland 题目连接: http://www.codeforces.com/contest/638/problem/B Description Berlan ...
随机推荐
- activity入门
1.前单文件 <activity android:name="com.example.twoactivity.OtherScreenActivity" android:lab ...
- 纠错记录(Could not open the editor: Android XML Editor cannot process this input.)
Eclipse无法打开xml文件 preference->general->edit->file association->file types->选择.xml,然后在a ...
- 【转】CentOS6.5 增加一个SFTP上传的用户
原文链接地址:http://www.msits.com/archives/4477.html #创建sftp组groupadd sftp#创建一个用户zjhpuseradd -g sftp -s /b ...
- 12-UIKit(View绘制、绘制曲线、绘制文字、贴图)
目录: 1. View绘制 2. 绘制曲线 3. 绘制文字 4. 贴图 回到顶部 1. View绘制 1.1 做出自己的视图对象 TRCell : UITableViewCell : UIView U ...
- 演练5-3:Contoso大学校园管理系统3
在前面的教程中,我们使用了一个简单的数据模型,包括三个数据实体.在这个教程汇中,我们将添加更多的实体和关系,按照特定的格式和验证规则等自定义数据模型. Contoso大学校园管理系统的数据模型如下. ...
- Homebrew 1.0.0 发布,MacOS 上的包管理器
神器,没有它不知道怎么用macos https://www.oschina.net/news/77367/homebrew-1-0-0
- Android 播放声音
public static void PlayAlarmRing(Context mContext) { Uri alert = RingtoneManager.getDefaultUri(Ringt ...
- docker学习笔记4:利用docker hub上的mysql镜像创建mysql容器
docker hub上有官方的mysql镜像,我们可以利用它来创建mysql容器,作为一个服务容器使用. 1.下载mysql镜像 docker pull mysql 2.创建镜像 docker run ...
- ACM一些题目
Low Power 先二分答案,可以通过调整证明同一台机器选的两个芯片必然是提供能量数值相邻的两个.所以再贪心一下就可以了. 时间复杂度\(O(n \log n)\). Factors 假设\(k\) ...
- Silk Icons —— 再来 700 个免费小图标
http://mp.weixin.qq.com/mp/appmsg/show?__biz=MjM5NzM0MjcyMQ==&appmsgid=10000977&itemidx=2&am ...