题目

链接

http://poj.org/problem?id=1028

字面描述

描述

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/

输入描述

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.

输出描述

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.

样例

输入样例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

输出样例1

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

思路

按照题目翻译的用2个栈去实现就完了。别说看不懂

代码实现

#include<iostream>
#include<string.h>
#include<cstring>
using namespace std;const int maxn=100+10;
int top1,top2;
string x,y,op;
string s1[maxn],s2[maxn];
int main(){op="http://www.acm.org/";while(1){cin>>x;if(x=="QUIT"){//cout<<"Ignored"<<endl;break;}else if(x=="VISIT"){cin>>y;s2[top2++]=op;op=y;top1=0;}else if(x=="BACK"){if(!top2){cout<<"Ignored"<<endl;continue;}s1[top1++]=op;op=s2[--top2];}else {if(!top1){cout<<"Ignored"<<endl;continue;}s2[top2++]=op;op=s1[--top1];}int len=op.length();for(int i=0;i<len;i++)printf("%c",op[i]);cout<<endl;}return 0;
}