题目链接:http://poj.org/problem?

id=1028

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

题意:

模拟浏览器浏览网页是前后翻页。

思路:

开两个栈。分别存储当前网页前后的网页,注意新訪问一个网页时,应该把当前网页的前面的栈清空!

代码例如以下:

#include <iostream>
#include <algorithm>
#include <string>
#include <stack>
using namespace std;
stack <string>b;
stack <string>f;
int main()
{
string tt = "http://www.acm.org/";
string s;
while(cin >> s)
{
if(s == "QUIT")
break;
if(s == "VISIT")
{
b.push(tt);
cin >> tt;
cout<<tt<<endl;//始终输出当前页
while(!f.empty())//当新訪问一个页面的时候把之前页面前面的清空
{
f.pop();
}
}
else if(s == "BACK")
{
if(!b.empty())
{
f.push(tt);
tt = b.top();
b.pop();
cout<<tt<<endl;//始终输出当前页
}
else
cout<<"Ignored"<<endl;
}
else
{
if(!f.empty())
{
b.push(tt);
tt = f.top();
f.pop();
cout<<tt<<endl;//始终输出当前页
}
else
cout<<"Ignored"<<endl;
}
}
return 0;
}

poj 1028 Web Navigation(模拟)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. poj 1028 Web Navigation 【模拟题】

    题目地址:http://poj.org/problem?id=1028 测试样例: Sample Input VISIT http://acm.ashland.edu/ VISIT http://ac ...

  3. 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 ...

  4. 1028 Web Navigation

    题目链接: http://poj.org/problem?id=1028 题意: 模拟浏览器的前进/后退/访问/退出 的四个操作. 输出当前访问的URL或者Ignore(如果不能前进/后退). 分析: ...

  5. poj 1208 Web Navigation(堆栈操作)

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

  6. POJ 1028:Web Navigation

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

  7. 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. ...

  8. 【stack】模拟网页浏览 poj 1028

    #include<stdio.h> #include<string.h> int main() { ][]; ]; int i,depth; strcpy(s[]," ...

  9. 使用 Web API 模拟其他用户

    模拟的要求 模拟可代表另一个 Microsoft Dynamics CRM 用户,用于执行业务逻辑(代码)以便提供所需功能或服务,它使用模拟用户的相应角色和基于对象的安全性.这项技术很有必要,因为 M ...

随机推荐

  1. 洛谷P1972 HH的项链

    传送门啦 分析: 题目描述不说了,大意是,求一段区间内不同元素的种数. 看到区间,我们大概先想到的是暴力(然后炸掉).线段树.树状数组.分块. 下面给出的是一种树状数组的想法. 首先,对于每一段区间里 ...

  2. Luogu P1750 【出栈序列】

    一眼(万年)贪心minn设小调不出来祭 首先要保证更靠前的输出更小那么容易想到,对于之后可能入栈的元素(即栈的剩余空间仍能装下的所有元素),我们可以取其中的最小值minn,和栈顶元素$top$比较,如 ...

  3. 网络编程--Socket与ServerSocket

    1.服务器端代码 package net; import java.io.PrintStream; import java.net.ServerSocket; import java.net.Sock ...

  4. 浅析redux

    一 redux 思想 首先,每一个webApp有且只有一个state tree,为方便管理和跟踪state的变化,也为了减少混乱,redux只允许通过发送(dispatch)action的方式来改变s ...

  5. JavaScript 三个常用对话框

    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/ ...

  6. mysql 5.1 下载地址 百度云网盘下载

    mysql 百度云下载 链接: https://pan.baidu.com/s/1fPSEcgtDN7aU2oQ_sL08Ww 提取码: 关注公众号[GitHubCN]回复2539获取

  7. Linux的bash快捷键

    Ctrl-A 相当于HOME键,用于将光标定位到本行最前面 Ctrl-E 相当于End键,即将光标移动到本行末尾 Ctrl-B 相当于左箭头键,用于将光标向左移动一格 Ctrl-F 相当于右箭头键,用 ...

  8. 【POJ】1286.Necklace of Beads

    题解 群论,我们只要找出所有的置换群的所有循环节 具体可参照算法艺术与信息学竞赛 旋转的置换有n个,每一个的循环节个数是gcd(N,i),i的范围是0到N - 1 翻转,对于奇数来说固定一个点,然后剩 ...

  9. thinkphp getField()获取一列或一个数据

    在开发中经常要获取一个数据的情况,thinkphp中有一个getField()方法可以解决这个问题. 获取一个数据 1 2 $user = M('demo'); $data = $user->g ...

  10. 有了这套flex页面布局方案,页面什么的,那都不是事。

    一.CSS3弹性盒子弹性盒子是CSS3的一种新布局模式.CSS3 弹性盒( Flexible Box 或 flexbox),是一种当页面需要适应不同的屏幕大小以及设备类型时确保元素拥有恰当的行为的布局 ...