poj 1028 Web Navigation

题目：

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/

 

源代码：

#include<iostream>#include<stack>#include<string>using namespace std;class URLvisit{public:URLvisit(){backwardSt.push("http://www.acm.org/");}void solution(void){string cmd,url;while((cin>>cmd)&&cmd!="QUIT"){if("VISIT" == cmd){cin>>url;backwardSt.push(url);cout<<url<<endl;while(!forwardSt.empty())forwardSt.pop();}if("BACK" == cmd) {if(backwardSt.size()>1){forwardSt.push(backwardSt.top());backwardSt.pop();cout<<backwardSt.top()<<endl;}else cout<<"Ignored"<<endl;}    if ("FORWARD" == cmd){if(!forwardSt.empty()){cout<<forwardSt.top()<<endl;backwardSt.push(forwardSt.top());forwardSt.pop();}else cout<<"Ignored"<<endl;}}};private:stack<string> backwardSt;stack<string> forwardSt;};int main(){URLvisit poj1028;poj1028.solution();system("pause");return 0;}