Problem 2656 --Elementary data structures - Queue

2656: Elementary data structures - Queue

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Time Limit $1$ 秒/Second(s) Memory Limit $512$ 兆字节/Megabyte(s)
提交总数 $18$ 正确数量 $12$
裁判形式 标准裁判/Standard Judge 我的状态 尚未尝试
难度 分类标签 STL

There are n processes in a queue. Each process has namei and timei. The round-robin scheduling handles the processes in order. A round-robin scheduler gives each process a quantum (a time slot) and interrupts the process if it is not completed by then. The process is resumed and moved to the end of the queue, then the scheduler handles the next process in the queue.

For example, we have the following queue with the quantum of 100ms.

A(150) - B(80) - C(200) - D(200)

First, process A is handled for 100ms, then the process is moved to the end of the queue with the remaining time (50ms). 

B(80) - C(200) - D(200) - A(50)

Next, process B is handled for 80ms. The process is completed with the time stamp of 180ms and removed from the queue. 

C(200) - D(200) - A(50) 

Your task is to write a program which simulates the round-robin scheduling.

n q

name1 time1 

name2 time2 

... 

namen  timen 

In the first line the number of processes n and the quantum q are given separated by a single space. 

In the following n lines, names and times for the n processes are given. namei and timei are separated by a single space.

For each process, prints its name and the time the process finished in order.
5 100
p1 150
p2 80
p3 200
p4 350
p5 20
p2 180
p5 400
p1 450
p3 550
p4 800

1 ≤  n ≤  100000

1 ≤  q ≤  1000 

1 ≤  timei ≤  50000

1 ≤  length of namei ≤  10

1 ≤  Sum of timei ≤  1000000

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本题记录 用 户(点击查看用户) 运行号(点击购买题解) 时 间
算法最快[$17 $ms] AOJ大管家 437169 2019-06-08 00:04:26
内存最少[$3088 $KB] beholder 779383 2021-11-14 23:30:42
第一AC 计爱玲 359115 2018-12-31 11:02:03
第一挑战 计爱玲 359115 2018-12-31 11:02:03

赛题来源/所属竞赛 会津大学《挑战数据结构与算法》 挑战数据结构与算法

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