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  • POJ1028 Web Navigation

    题目来源:http://poj.org/problem?id=1028

    题目大意:

      模拟实现一个浏览器的“前进”和“回退”功能。由一个forward stack和一个backward stack实现。

      打开浏览器时的正位于http://www.acm.org/。然后浏览器会接受下面四种命令:

      BACK:将当前页面压入forward stack,将backward stack顶部页面弹出,成为当前页面。若当前backward stack为空,忽略该命令;

      FORWARD:将当前页面压入backward stack,将forward stack顶部的页面弹出,作为当前页面。若当前forward stack为空,忽略该命令;

      VISIT:将当前页面压入backward stack,将指定的URL作为当前页面;

      QUIT:退出浏览器。

    输入:命令的序列,以QUIT作为结束。

    输出:如果当前页面发生了改变,输出改变后的当前页面,如果命令被忽略,输出“Ignored.”


    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

    题干已经把如何实现讲解得很清楚,照着写就可以了,要是不给提示,这倒会是道很有意思的题。

     1 //////////////////////////////////////////////////////////////////////////
     2 //        POJ1028 Web Navigation
     3 //        Memory: 264K        Time: 47MS
     4 //        Language: C++        Result: Accepted
     5 //////////////////////////////////////////////////////////////////////////
     6 
     7 #include <iostream>
     8 #include <string>
     9 #include <stack>
    10 
    11 using namespace std;
    12 
    13 stack<string> forwardStack;
    14 stack<string> backwardStack; 
    15 
    16 void clearForwardStack() {
    17     while (!forwardStack.empty()) {
    18         forwardStack.pop();
    19     }
    20 }
    21 
    22 int main() {
    23     string operation;
    24     string currentURL = "http://www.acm.org/";
    25     string nextURL;
    26     while (cin >> operation && operation != "QUIT") {
    27 
    28         switch (operation[0]) {
    29             case 'V':
    30                 cin >> nextURL;
    31                 backwardStack.push(currentURL);
    32                 currentURL = nextURL;
    33                 cout << currentURL << endl;
    34                 clearForwardStack();
    35                 break;
    36             case 'B':
    37                 if (backwardStack.empty()) {
    38                     cout << "Ignored" << endl;
    39                 } else {
    40                     forwardStack.push(currentURL);
    41                     currentURL = backwardStack.top();
    42                     cout << currentURL << endl;
    43                     backwardStack.pop();
    44                 }
    45                 break;
    46             case 'F':
    47                 if (forwardStack.empty()) {
    48                     cout << "Ignored" << endl;
    49                 } else {
    50                     backwardStack.push(currentURL);
    51                     currentURL = forwardStack.top();
    52                     cout << currentURL << endl;
    53                     forwardStack.pop();
    54                 }
    55                 break;
    56         }
    57     }
    58     system("pause");
    59     return 0;
    60 }
    View Code
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  • 原文地址:https://www.cnblogs.com/dengeven/p/3228997.html
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