Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 34314 | Accepted: 15322 |
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
问题链接:POJ1028 Web Navigation。
题意简述:(略)
问题分析:这个题的原题应该是UVALive2356,只是输入格式略有不同。这个是有关浏览器操作的问题,直接模拟。
程序说明:需要使用两个堆栈存储访问履历,以备操作使用。使用堆栈是因为操作过程与堆栈是相似的。
参考链接:UVALive2356 ZOJ1061 Web Navigation【堆栈+模拟】
AC的C++语言程序如下:
/* POJ1028 Web Navigation */ #include <iostream> #include <string> #include <stack> using namespace std; int main() { stack<string> ss, st; string cmd, url; ss.push("http://www.acm.org/"); while(cin >> cmd) { if(cmd[0] == 'Q') // QUIT break; else if(cmd[0] == 'V') { // VISIT cin >> url; ss.push(url); cout << url << endl; // 清空:一旦输入一个新的URL,就不能再做FORWARD了 while(!st.empty()) st.pop(); } else if(cmd[0] == 'B') { // BACK if(ss.size() > 1) { st.push(ss.top()); ss.pop(); cout << ss.top() << endl; } else cout << "Ignored" << endl; } else if(cmd[0] == 'F') { // FORWARD if(!st.empty()) { ss.push(st.top()); cout << st.top() << endl; st.pop(); } else cout << "Ignored" << endl; } } return 0; }