zoukankan      html  css  js  c++  java
  • Codeforces Round #590 (Div. 3) B2. Social Network (hard version)

    链接:

    https://codeforces.com/contest/1234/problem/B2

    题意:

    The only difference between easy and hard versions are constraints on n and k.

    You are messaging in one of the popular social networks via your smartphone. Your smartphone can show at most k most recent conversations with your friends. Initially, the screen is empty (i.e. the number of displayed conversations equals 0).

    Each conversation is between you and some of your friends. There is at most one conversation with any of your friends. So each conversation is uniquely defined by your friend.

    You (suddenly!) have the ability to see the future. You know that during the day you will receive n messages, the i-th message will be received from the friend with ID idi (1≤idi≤109).

    If you receive a message from idi in the conversation which is currently displayed on the smartphone then nothing happens: the conversations of the screen do not change and do not change their order, you read the message and continue waiting for new messages.

    Otherwise (i.e. if there is no conversation with idi on the screen):

    Firstly, if the number of conversations displayed on the screen is k, the last conversation (which has the position k) is removed from the screen.
    Now the number of conversations on the screen is guaranteed to be less than k and the conversation with the friend idi is not displayed on the screen.
    The conversation with the friend idi appears on the first (the topmost) position on the screen and all the other displayed conversations are shifted one position down.
    Your task is to find the list of conversations (in the order they are displayed on the screen) after processing all n messages.

    思路:

    队列维护, map记录, 模拟即可.

    代码:

    #include <bits/stdc++.h>
    using namespace std;
    
    map<int, bool> Mp;
    deque<int> Que;
    int n, k;
    
    int main()
    {
        int id;
        scanf("%d%d", &n, &k);
        for (int i = 1;i <= n;i++)
        {
            scanf("%d", &id);
            if (!Mp[id])
            {
                Que.push_front(id);
                Mp[id] = true;
            }
            if (Que.size() > k)
            {
                int tmp = Que.back();
                Que.pop_back();
                Mp[tmp] = false;
            }
        }
        printf("%d
    ", (int)Que.size());
        while (!Que.empty())
        {
            printf("%d ", Que.front());
            Que.pop_front();
        }
    
        return 0;
    }
    
  • 相关阅读:
    巨人的崛起 Android操作系统发展历程
    iPhone 4S定位功能加强 支持GLONASS系统
    android子线程更新UI,与主Thread一起工作
    什么在阻碍移动互联网的普及?
    WEB.CONFIG 配置详解
    解决在aspx页面上进行传中文参数时会出现乱码问题
    VS2005水晶报表注册码
    windows2003应用程序池假死的问题
    浅谈ASP.NET内部机制
    合理建立数据库索引
  • 原文地址:https://www.cnblogs.com/YDDDD/p/11618493.html
Copyright © 2011-2022 走看看