Description

Standard web browsers contain features to move backward and forward among the pages recently visited. One way to implement these features is to use two stacks to keep track of the pages that can be reached by moving backward and forward. In this problem, you are asked to implement this. 
The following commands need to be supported: 
BACK: Push the current page on the top of the forward stack. Pop the page from the top of the backward stack, making it the new current page. If the backward stack is empty, the command is ignored. 
FORWARD: Push the current page on the top of the backward stack. Pop the page from the top of the forward stack, making it the new current page. If the forward stack is empty, the command is ignored. 
VISIT : Push the current page on the top of the backward stack, and make the URL specified the new current page. The forward stack is emptied. 
QUIT: Quit the browser. 
Assume that the browser initially loads the web page at the URL http://www.acm.org/

Input

Input is a sequence of commands. The command keywords BACK, FORWARD, VISIT, and QUIT are all in uppercase. URLs have no whitespace and have at most 70 characters. You may assume that no problem instance requires more than 100 elements in each stack at any time. The end of input is indicated by the QUIT command.

Output

For each command other than QUIT, print the URL of the current page after the command is executed if the command is not ignored. Otherwise, print "Ignored". The output for each command should be printed on its own line. No output is produced for the QUIT command.

Sample Input

VISIT http://acm.ashland.edu/
VISIT http://acm.baylor.edu/acmicpc/
BACK
BACK
BACK
FORWARD
VISIT http://www.ibm.com/
BACK
BACK
FORWARD
FORWARD
FORWARD
QUIT

Sample Output

http://acm.ashland.edu/
http://acm.baylor.edu/acmicpc/
http://acm.ashland.edu/
http://www.acm.org/
Ignored
http://acm.ashland.edu/
http://www.ibm.com/
http://acm.ashland.edu/
http://www.acm.org/
http://acm.ashland.edu/
http://www.ibm.com/
Ignored
 Source Code

 Problem:         User:
Memory: 200K Time: 0MS
Language: C++ Result: Accepted
Source Code
#include"iostream"
#include"cstring"
#include"string"
#include"stack"
using namespace std;
int main()
{
stack<string> s1,s2;
string s,ss;
s="http://www.acm.org/";
s1.push(s);
while(cin>>s)
{
if(s=="QUIT")
break;
if(s=="VISIT")
{
cin>>ss;
cout<<ss<<endl;
s1.push(ss);
while(!s2.empty())
s2.pop();
}
else if(s=="BACK")
{
//cout<<"back "<<s1.size()<<endl;
if(s1.size()>=)
{
s2.push(s1.top());
s1.pop();
cout<<s1.top()<<endl;
}
else
cout<<"Ignored"<<endl;
}
else if(s=="FORWARD")
{
if(!s2.empty())
{
cout<<s2.top()<<endl;
s1.push(s2.top());
s2.pop();
}
else
cout<<"Ignored"<<endl;
}
}
return ;
}

Web Navigation的更多相关文章

  1. poj 1028 Web Navigation

    Web Navigation Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 31088   Accepted: 13933 ...

  2. POJ1028 Web Navigation

    题目来源:http://poj.org/problem?id=1028 题目大意: 模拟实现一个浏览器的“前进”和“回退”功能.由一个forward stack和一个backward stack实现. ...

  3. POJ 1028:Web Navigation

    Web Navigation Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 30828   Accepted: 13821 ...

  4. POJ-1028 Web Navigation 和TOJ 1196. Web Navigation

    Standard web browsers contain features to move backward and forward among the pages recently visited ...

  5. [POJ1028]Web Navigation(栈)

    这题是01年East Central North的A题,目测是签到题 Description Standard web browsers contain features to move backwa ...

  6. poj 1028 Web Navigation(模拟)

    题目链接:http://poj.org/problem? id=1028 Description Standard web browsers contain features to move back ...

  7. poj 1208 Web Navigation(堆栈操作)

    一.Description Standard web browsers contain features to move backward and forward among the pages re ...

  8. ZOJ 1061 Web Navigation

    原题链接 题目大意:模拟一个浏览器,打开一个网页.后退或者前进,输出网址. 解法:用两个堆栈分别表示后退保存的网页和前进保存的网页.初始化时把当前页面压入后退堆栈的栈顶.要注意几点,一个是每次记得要清 ...

  9. POJ 1028 Web Navigation 题解

    考查代码能力的题目.也能够说是算法水题,呵呵. 推荐新手练习代码能力. 要添加难度就使用纯C实现一下stack,那么就有点难度了,能够使用数组模拟环形栈.做多了,我就直接使用STL了. #includ ...

随机推荐

  1. centos编译helloworld的几个小问题

    1.GCC使用在使用GCC编译程序时,编译过程可以被细分为四个阶段:预处理(Pre-Processing)编译(Compiling)汇编(Assembling)链接(Linking).例如:      ...

  2. 非常实用的Ubuntu常用终端命令

    先介绍关于文件和目录的命令: ls 列出当前目录文件(不包括隐含文件) ls -a 列出当前目录文件(包括隐含文件) ls -l 列出当前目录下文件的详细信息 cd .. 回当前目录的上一级目录 cd ...

  3. linux socket中的SO_REUSEADDR

    Welcome to the wonderful world of portability... or rather the lack of it. Before we start analyzing ...

  4. mount失败

    又一次遇到mount失败,提示文件系统类型错误.选项错误.有坏超级块等.之前是在ubuntu 14.04 lts desktop上挂载windows下共享文件夹遇到的.这次具体的环境如下:CentOS ...

  5. C++问题-UniqueAppObject.cpp(147): error C3861: “GUXClientInit”: 找不到标识符

    问题经过:在同事的产品上增加新功能,拿来的代码包,用VS打开后,提示某个文件不存在,从项目中移除.CPP.H文件后,提示错误,提示如下:1>UniqueAppObject.cpp(147): e ...

  6. HDU 4562 守护雅典娜(dp)

    守护雅典娜 Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submi ...

  7. 查杀oracle锁表

    ()锁表查询的代码有以下的形式: select count(*) from v$locked_object; select * from v$locked_object; ()查看哪个表被锁 sele ...

  8. [HDU 4089]Activation[概率DP]

    题意: 有n个人排队等着在官网上激活游戏.Tomato排在第m个. 对于队列中的第一个人.有以下情况: 1.激活失败,留在队列中等待下一次激活(概率为p1) 2.失去连接,出队列,然后排在队列的最后( ...

  9. Codeforces Round #260 (Div. 1) A. Boredom (简单dp)

    题目链接:http://codeforces.com/problemset/problem/455/A 给你n个数,要是其中取一个大小为x的数,那x+1和x-1都不能取了,问你最后取完最大的和是多少. ...

  10. Educational Codeforces Round 10 D. Nested Segments (树状数组)

    题目链接:http://codeforces.com/problemset/problem/652/D 给你n个不同的区间,L或者R不会出现相同的数字,问你每一个区间包含多少个区间. 我是先把每个区间 ...