POJ 1028 用栈模拟浏览器的前进与后退
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poj 1028
Web Navigation
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 17718 Accepted: 7751
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/BACKBACKBACKFORWARDVISIT http://www.ibm.com/BACKBACKFORWARDFORWARDFORWARDQUIT
Sample Output
http://acm.ashland.edu/http://acm.baylor.edu/acmicpc/http://acm.ashland.edu/http://www.acm.org/Ignoredhttp://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
East Central North America 2001
my AC code
- #pragma warning (disable: 4786)
- #include <iostream>
- #include <stack>
- #include <string>
- using namespace std;
- //要注意当前访问的页面就是backset栈顶的元素,切记
- //理解浏览器前进后退的含义自己编
- int main(){
- string brower;
- stack<string> backst,forst;
- string command,url;
- backst.push("http://www.acm.org/");
- while (cin>>command,command!="QUIT")
- {
- if (command == "VISIT")
- {
- cin>>url;
- backst.push(url);
- cout<<url<<endl;
- while(!forst.empty()) forst.pop();
- }
- else if (command == "BACK")
- {
- if (backst.size()==1)//等于1表示就是当前的页面,已经不能够弹栈了
- {
- cout<<"Ignored"<<endl;
- }
- else
- {
- url = backst.top();
- forst.push(url);
- backst.pop();
- url = backst.top();
- cout<<url<<endl;
- }
- }
- else if (command == "FORWARD")
- {
- if (forst.size()==0)//forst的大小可以为零,这点要与backset区分开来
- {
- cout<<"Ignored"<<endl;
- }
- else
- {
- url= forst.top();
- backst.push(url);
- forst.pop();
- cout<<url<<endl;
- }
- }
- }
- return 0;
- }
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