poj-1028 -网页导航

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 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.

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.

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

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
题目大意：

        说的是栈的模拟，但是我还不能很好的理解运用栈，所以我就这样做了，调试了好半天
#include<iostream>
#include<string>
using namespace std;
int main()
{
   string str[100];
    string s1="http://www.acm.org/",s2;
str[0]=s1;
int q=0,ed=0;
    while(1)
    {cin>>s1;
        if(s1=="QUIT"){break;}

        if(s1=="VISIT")
        {
        cin>>s2;
    q++;
      str[q]=s2;
      cout<<str[q]<<endl;
ed=q;    //每次添加新的url时都要让ed=q；q相当于一直指向当前所指的url下标； 而ed是最后一个输入的url下标。
        }
        if(s1=="BACK")
        {if(q<=0){ cout<<"Ignored"<<endl;}//若把if 和 else 的条件互换，过程会变得更加繁琐；
          else {cout<<str[--q]<<endl;}  

        }
        if(s1=="FORWARD")
        {
         if(q>=ed){  cout<<"Ignored"<<endl;}
            else {cout<<str[++q]<<endl;}

        }

    }


    return 0;

和我思路一样的代码，总觉的人家的简单清晰很多
并且我最后一个bug还是看这个代码找到的

#include<stdio.h>
 char str[200][71]={"http://www.acm.org/"};
int point=0,end=0;
void forword()
{
    if(point>=end) printf("Ignored\n");//我在参考这儿的写法
    else printf("%s\n",str[++point]);
}
void back()
{
    if(point<=0) printf("Ignored\n");
    else printf("%s\n",str[--point]);
}
void vist()
{
    scanf("%s",str[++point]);
    printf("%s\n",str[point]);
    end=point;                                 //在此处更新栈的最终指针
}
int main()
{
    char com[10];
    int g=1;
    while(g)
    {
        scanf("%s",com);
        switch(com[0])
        {
            case ‘V‘:vist();break;
            case ‘B‘:back();break;
            case ‘F‘:forword();break;
            default : g=0;break;
        }
    }
    return 0;
}

}