UVALive2356 ZOJ1061 Web Navigation【堆栈+模拟】

Web Navigation

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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 <url>: 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/

This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

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

1

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

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Source: East Central North America 2001

Regionals 2001 >> North America - East Central NA

问题链接：UVALive2356 ZOJ1061 Web Navigation。

题意简述：（略）

问题分析：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。

AC的C++语言程序如下：

/* UVALive2356 ZOJ1061 Web Navigation */

#include <iostream>
#include <string>
#include <stack>

using namespace std;

int main()
{
    int n;

    cin >> n;
    while(n--) {
        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;
            }
        }

        if(n)
            cout << endl;
    }

    return 0;
}