Web Navigation

Description

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

这道题就是对网页实际的操作，前进和后退，理解题目就行，用两个栈，一个保存前进，一个保存后退就行了。

#include <stdio.h>
#include <string.h>
const int maxn = 75, N = 10005;
char q[N][maxn], queue[N][maxn], str[maxn];
char ch[maxn];
int main ( )
{
    int T, top, front, cas = 0;
    //freopen ( "in0.in", "r", stdin );
    scanf ( "%d", &T );
    while ( T -- )
    {
        front = top = -1;
        strcpy ( q[++ top], "http://www.acm.org/" );
        //开始就将一个网址加入栈
        if ( cas ++ )   //每组数据有换行
            printf ( "\n" );
        while ( ~ scanf ( "%s", str ) && strcmp ( str, "QUIT" ) != 0 )
        {
            if ( strcmp ( str, "VISIT" ) == 0 )
            {   //查看时将所有可以后退的全部去掉
                scanf ( "%s", ch );
                printf ( "%s\n", ch );
                strcpy ( q[++ top], ch );
                front = -1;
            }
            else if ( strcmp ( str, "BACK" ) == 0 )
            {
                if ( top <= 0 ) //小于等于0证明不能在后退了
                    printf ( "Ignored\n" );
                else
                {
                    //将此网页加入可以前进的栈中
                    strcpy ( queue[++ front], q[top] );
                    printf ( "%s\n", q[-- top] );
                }
            }
            else
            {
                if ( front < 0 )
                    printf ( "Ignored\n" );
                else
                {
                    //将前进的网页加到后退中
                    strcpy ( q[++ top], queue[front] );
                    printf ( "%s\n", queue[front --] );
                }
            }
        }
    }
    return 0;
}