codeforces644B. Processing Queries (模拟)
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.
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.
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.
5 1
2 9
4 8
10 9
15 2
19 1
11 19 -1 21 22
4 1
2 8
4 8
10 9
15 2
10 18 27 -1
Consider the first sample.
- The server will start to process first query at the moment 2 and will finish to process it at the moment 11.
- At the moment 4 second query appears and proceeds to the queue.
- At the moment 10 third query appears. However, the server is still busy with query 1, b = 1 and
there is already query 2 pending in the queue, so third query is just rejected. - At the moment 11 server will finish to process first query and will take the second query from the queue.
- At the moment 15 fourth query appears. As the server is currently busy it proceeds to the queue.
- At the moment 19 two events occur simultaneously: server finishes to proceed the second query and the fifth query appears. As was said in the
statement above, first server will finish to process the second query, then it will pick the fourth query from the queue and only then will the fifth query appear. As the queue is empty fifth query is proceed there. - Server finishes to process query number 4 at the moment 21.
Query number 5 is picked from the queue. - Server finishes to process query number 5 at the moment 22.
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<bitset>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef long double ldb;
#define inf 99999999
#define pi acos(-1.0)
#define maxn 200050
ll t[maxn],d[maxn];
int q[511111];
ll ans[maxn];
int main()
{
int i,j,x;
ll n,m,b,ed;
while(scanf("%lld%lld",&n,&b)!=EOF)
{
for(i=1;i<=n;i++){
scanf("%lld%lld",&t[i],&d[i]);
}
int front,rear;
front=1;rear=0;
ed=0;
for(i=1;i<=n;i++){
while(front<=rear){
x=q[front];
if(ed<=t[i]){
ed=max(ed,t[x])+d[x];
ans[x]=ed;
front++;
}
else break;
}
if(ed<=t[i] && front>rear ){
ed=t[i]+d[i];
ans[i]=ed;
}
else if(rear-front+1<b){
rear++;
q[rear]=i;
}
else ans[i]=-1;
}
while(front<=rear){
x=q[front];
front++;
ed=max(ed,t[x])+d[x];
ans[x]=ed;
}
for(i=1;i<=n;i++){
if(i==n)printf("%lld\n",ans[i]);
else printf("%lld ",ans[i]);
}
}
return 0;
}
codeforces644B. Processing Queries (模拟)的更多相关文章
- CROC 2016 - Qualification B. Processing Queries 模拟
B. Processing Queries 题目连接: http://www.codeforces.com/contest/644/problem/B Description In this prob ...
- Code Forces 644B Processing Queries
B. Processing Queries time limit per test5 seconds memory limit per test256 megabytes inputstandard ...
- 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 ...
- 代码本色 用编程模拟自然系统 (Daniel Shiffman 著)
https://cdnjs.cloudflare.com/ajax/libs/p5.js/0.5.7/p5.js http://www.box2d.org http://www.jbox2d.org ...
- Programming Entity Framework 翻译(1)-目录
1. Introducing the ADO.NET Entity Framework ado.net entity framework 介绍 1 The Entity Relationship Mo ...
- 【mysql】关于临时表
mysql官方的介绍 In some cases, the server creates internal temporary tables while processing queries. Suc ...
- open_table与opened_table
好多人在调优Mysql的时候,总是对open_tables和opend_tables两个参数分别不清. 网上好多解释都是这样的:open_tables:当前打开表的数量opened_tables:当前 ...
- 【MySQL】查询使用临时表
MySQL查询产生临时表的分析 官网说明的地址:http://dev.mysql.com/doc/refman/5.5/en/internal-temporary-tables.html 参考:htt ...
- mysql --The MEMORY Storage Engine--官方文档
原文地址:http://dev.mysql.com/doc/refman/5.7/en/memory-storage-engine.html The MEMORY storage engine (fo ...
随机推荐
- Shiro的认证与授权
shiro实战教程 一.权限管理 1.1什么是权限管理 基本上涉及到用户参与的系统都需要进行权限管理,权限管理属于系统安全的范畴,权限管理实现对用户访问系统的控制,按照安全规则或者安全策略控制用户可以 ...
- Java多线程-锁的区别与使用
目录 锁类型 可中断锁 公平锁/非公平锁 可重入锁 独享锁/共享锁 互斥锁/读写锁 乐观锁/悲观锁 分段锁 偏向锁/轻量级锁/重量级锁 自旋锁 Synchronized与Static Synchron ...
- 如何构建一个多人(.io) Web 游戏,第 1 部分
原文:How to Build a Multiplayer (.io) Web Game, Part 1 GitHub: https://github.com/vzhou842/example-.io ...
- 基于Python开发数据宽表实例
搭建宽表作用,就是为了让业务部门的数据分析人员,在日常工作可以直接提取所需指标,快速做出对应专题的数据分析.在实际工作中,数据量及数据源繁多,如果每个数据分析人员都从计算加工到出报告,除了工作效率巨慢 ...
- 系统吞吐量与QPS/TPS
QPS/TPS QPS:Queries Per Second意思是"每秒查询率",是一台服务器每秒能够相应的查询次数,是对一个特定的查询服务器在规定时间内所处理流量多少的衡量标准. ...
- 【Java】Java注释 - 单行、块、文档注释
简单记录,Java 核心技术卷I 基础知识(原书第10 版) 注释 我们在编写程序时,经常需要添加一些注释,用来描述某段代码的作用,提高Java源程序代码的可读性,使得Java程序条理清晰. 写代码的 ...
- Python 日志打印之自定义logger handler
日志打印之自定义logger handler By:授客 QQ:1033553122 #实践环境 WIN 10 Python 3.6.5 #实践代码 handler.py #!/usr/bin/env ...
- 控制tomcat日志文件的输出到catalina.out
在catalina.sh中直接把下面的内容注释掉即可:
- Maven 依赖机制
概述 在 Maven 依赖机制的帮助下自动下载所有必需的依赖库,并保持版本升级.让我们看一个案例研究,以了解它是如何工作的.假设你想使用 Log4j 作为项目的日志.这里你要做什么? 传统方式 访问 ...
- 日志采集技术分析 Inode Inotify
日志采集技术分析[阿里] - 新手学习导向 - 中国红客联盟 - Powered by HUC http://www.cnhonkerarmy.com/thread-236973-1-1.html