zoukankan      html  css  js  c++  java
  • CDZSC_2015寒假新人(4)——搜索 N

    Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

    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 : 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

    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

    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




    思路:这题模拟栈。自己开一个数组和栈头指针和栈尾指针来模拟。。。。。坑点:因为浏览器的特点当你在前进后退的途中输入一个网站,那么他那么这个网站的位置到栈尾的网站全部清空了,也就是说栈尾就是你输入的这个网站。



    #include <iostream>
    #include <cstdio>
    #include <cstring>
    using namespace std;
    char str[100][130];
    int main()
    {
    #ifdef CDZSC_OFFLINE
        freopen("in.txt","r",stdin);
    #endif
        char str1[10],a[100];
        int n,m;
        n=0,m=0;
        memset(str,0,sizeof(str));
        strcpy(str[0],"http://www.acm.org/");
        while(scanf("%s",str1)!=EOF)
        {
            if(strcmp(str1,"VISIT")==0)
            {
                scanf("%s",a);
                strcpy(str[++n],a);
                m=n;//后面的网站会被清空
                printf("%s
    ",str[m]);
            }
            else if(strcmp(str1,"BACK")==0)
            {
                n--;
                if(n<0)
                {
                    n=0;
                    printf("Ignored
    ");
                }
                else
                {
                    printf("%s
    ",str[n]);
                }
            }
            else if(strcmp(str1,"FORWARD")==0)
            {
                n++;
                if(n>m)
                {
                    n=m;
                    printf("Ignored
    ");
                }
                else
                {
                    printf("%s
    ",str[n]);
                }
            }
            else if(strcmp(str1,"QUIT")==0)
            {
                break;
            }
        }
        return 0;
    }
  • 相关阅读:
    log&& buffevent&&内存池 1
    ngx内存池设计概阅
    读 perf 笔记 简写
    smaps 使用&& 内存泄露
    cache占用高 文件delete cache
    工具小用法 dropwatch ss perf
    golang 读书笔记 数据类型
    重看ebpf 通信&&数据结构分析
    TCP 发送缓冲区问题--根本原因是gso引起 转载
    重看ebpf -代码载入执行点-hook
  • 原文地址:https://www.cnblogs.com/Wing0624/p/4253954.html
Copyright © 2011-2022 走看看