POJ1028 Web Navigation
2012-11-15 21:06
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Web Navigation
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
Sample Output
Source
East Central North America 2001
这是一道模拟网页浏览的题目,模拟访问网站,前进后退等。可以用类似链表的方法来储存网页间的前进后退关系。
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 24561 | Accepted: 10944 |
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
East Central North America 2001
这是一道模拟网页浏览的题目,模拟访问网站,前进后退等。可以用类似链表的方法来储存网页间的前进后退关系。
#include<iostream> #include<cstring> #include<cstdio> using namespace std; struct node { int back;//储存后退 int forward;//储存前进 char c[100];//网址 }p[150]; int main() { for(int i=1;i<150;i++) { p[i].back=0;//默认0为没有 p[i].forward=0; } int v=1; int index=1; strcpy(p[1].c,”http://www.acm.org/”);//首页 string a; char b[100]; while(cin>>a) { if(a==”QUIT”)return 0; if(a==”BACK”) { if(p[index].back==0)cout<<”Ignored”<<endl;//默认的0为没有后退项 else {//此处更改链表 cout<<p[p[index].back].c<<endl; p[p[index].back].forward=index; index=p[index].back; } } if(a==”FORWARD”) { if(p[index].forward==0)cout<<”Ignored”<<endl;//0为没有前进项 else {//更改链表 cout<<p[p[index].forward].c<<endl; p[p[index].forward].back=index; index=p[index].forward; } } if(a==”VISIT”) { scanf(“%s”,&p[++v].c); cout<<p[v].c<<endl; p[v].back=index; index=v; } } return 0; }
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