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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/
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;
}
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