B. Processing Queries

题目连接:

http://www.codeforces.com/contest/644/problem/B

Description

In this problem you have to simulate the workflow of one-thread server. There are n queries to process, the i-th will be received at moment ti and needs to be processed for di units of time. All ti are guaranteed to be distinct.

When a query appears server may react in three possible ways:

If server is free and query queue is empty, then server immediately starts to process this query.

If server is busy and there are less than b queries in the queue, then new query is added to the end of the queue.

If server is busy and there are already b queries pending in the queue, then new query is just rejected and will never be processed.

As soon as server finished to process some query, it picks new one from the queue (if it's not empty, of course). If a new query comes at some moment x, and the server finishes to process another query at exactly the same moment, we consider that first query is picked from the queue and only then new query appears.

For each query find the moment when the server will finish to process it or print -1 if this query will be rejected.

Input

The first line of the input contains two integers n and b (1 ≤ n, b ≤ 200 000) — the number of queries and the maximum possible size of the query queue.

Then follow n lines with queries descriptions (in chronological order). Each description consists of two integers ti and di (1 ≤ ti, di ≤ 109), where ti is the moment of time when the i-th query appears and di is the time server needs to process it. It is guaranteed that ti - 1 < ti for all i > 1.

Output

Print the sequence of n integers e1, e2, ..., en, where ei is the moment the server will finish to process the i-th query (queries are numbered in the order they appear in the input) or  - 1 if the corresponding query will be rejected.

Sample Input

5 1

2 9

4 8

10 9

15 2

19 1

Sample Output

11 19 -1 21 22

Hint

题意

有一个任务队列,你队列里面最多堆b个任务,如果新加入的任务的时候,任务队列已经满了的话,那这个任务就会被拒绝

现在给你n个任务的加入时间和处理时间

问你每个任务的完成情况是怎么样的。

题解:

模拟一下队列就好了

用一个类似two point的思想去处理每一个任务。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6+7; long long a[maxn];
int n,b;
long long ans[maxn];
int main()
{
scanf("%d%d",&n,&b);
int st=0,ed=0;
for(int i=1;i<=n;i++)
{
int x,y;scanf("%d%d",&x,&y);
while(st<ed&&a[st]<=x)st++;
if(st==ed)ans[i]=x+y,a[ed++]=ans[i];
else if(ed-st<=b)ans[i]=a[ed-1]+y,a[ed++]=ans[i];
else ans[i]=-1;
}
for(int i=1;i<=n;i++)
cout<<ans[i]<<" ";
cout<<endl;
}

CROC 2016 - Qualification B. Processing Queries 模拟的更多相关文章

  1. codeforces644B. Processing Queries (模拟)

    In this problem you have to simulate the workflow of one-thread server. There are n queries to proce ...

  2. CROC 2016 - Qualification C. Hostname Aliases map

    C. Hostname Aliases 题目连接: http://www.codeforces.com/contest/644/problem/C Description There are some ...

  3. Schlumberger Petrel 2016.3 地震解释 油藏模拟

    Schlumberger Petrel 2016.3 地震解释 油藏模拟世界上顶尖的三维地质建模软件,软件为用户提供的工具可以用于地震解释.地质建模.油藏数 值模拟等方面的使用,清晰的地质模型可以描述 ...

  4. Code Forces 644B Processing Queries

    B. Processing Queries time limit per test5 seconds memory limit per test256 megabytes inputstandard ...

  5. CROC 2016 - Elimination Round (Rated Unofficial Edition) D. Robot Rapping Results Report 二分+拓扑排序

    D. Robot Rapping Results Report 题目连接: http://www.codeforces.com/contest/655/problem/D Description Wh ...

  6. 2016年 实验四  B2B模拟实验

    实验四  B2B模拟实验 [实验目的] ⑴.掌握B2B中供应商的供求信息发布.阿里商铺开设和订单交易等过程. ⑵.掌握B2B中采购商的采购信息的发布.交易洽谈.网上支付和收货等过程. [实验条件] ⑴ ...

  7. 2016年 实验三 B2C模拟实验

    实验三 B2C模拟实验 [实验目的] 掌握网上购物的基本流程和B2C平台的运营 [实验条件] ⑴.个人计算机一台 ⑵.计算机通过局域网形式接入互联网. (3).奥派电子商务应用软件 [知识准备] 本实 ...

  8. Codeforces Round #515 (Div. 3) C. Books Queries (模拟)

    题意:有一个一维的书架,\(L\)表示在最左端放一本书,\(R\)表示在最右端放一本书,\(?\)表示从左数或从右数,最少数多少次才能得到要找的书. 题解:我们开一个稍微大一点的数组,从它的中间开始模 ...

  9. 2016.11.6 night NOIP模拟赛 考试整理

    题目+数据:链接:http://pan.baidu.com/s/1hssN8GG 密码:bjw8总结: 总分:300分,仅仅拿了120份. 这次所犯的失误:对于2,3题目,我刚刚看就想到了正确思路,急 ...

随机推荐

  1. android Timer TimerTask用法笔记

    Android中经常会遇到执行一些周期性定时执行的任务.初学的时候经常会使用Thread.sleep()方法.在android中,有Timer可以专门干这个事情. 先看看Timer.class中都是些 ...

  2. 第三讲:ifconfig:最熟悉又陌生的命令行

    你知道怎么查看IP地址吗? 当面试听到这个问题的时候,面试者常常会觉得走错了房间.我面试的是技术岗位啊,怎么问这么简单的问题? 的确,即便没有专业学过计算机的人,只要倒腾过电脑,重装过系统,大多也会知 ...

  3. 超简便安装mysql

    CentOS7默认数据库是mariadb,配置等用着不习惯,因此决定改成mysql,但是CentOS7的yum源中默认好像是没有mysql的.为了解决这个问题,我们要先下载mysql的repo源. 1 ...

  4. 宋牧春: Linux设备树文件结构与解析深度分析(1) 【转】

    转自:https://mp.weixin.qq.com/s/OX-aXd5MYlE_YoZ3p32qWA 作者简介 宋牧春,linux内核爱好者,喜欢阅读各种开源代码(uboot.linux.ucos ...

  5. WebBrowser中运行js

    HtmlElement script = wf.WebBrowser.Document.CreateElement("script"); script.SetAttribute(& ...

  6. 修改 firefox accesskey 的快捷键

    Chrome中,如果设置了 accesskey 的话,可以通过 Alt + 快捷键 来之直接跳转的.但在Firefox 中,可能是为了防止于菜单的快捷键冲突,所以设置了 Shift + Alt + 快 ...

  7. UTF-8和GB2312互转的最简单快捷的方法

    一.如果你想把utf-8转为GB2312 1.用记事本打开源码,把<meta http-equiv="Content-Type" content="text/htm ...

  8. Effective C++学习进阶版

    记得前段时间又一次拿起<Effective C++>的时候,有种豁然开朗的感觉,所以翻出了我第一遍读时做的笔记.只做参考以及查阅之用.如有需要请参阅<Effective C++> ...

  9. Merge k Sorted Lists——分治与堆排序(需要好好看)

    Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 1 ...

  10. thinkphp5.0目录结构

    下载最新版框架后,解压缩到web目录下面,可以看到初始的目录结构如下: project 应用部署目录 ├─application 应用目录(可设置) │ ├─common 公共模块目录(可更改) │ ...