Queue Aizu - ALDS1_3_B
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 .
Input
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.
Output
For each process, prints its name and the time the process finished in order.
Constraints
1 ≤ n ≤ 100000
1 ≤ q ≤ 1000
1 ≤ timei ≤ 50000
1 ≤ length of namei ≤ 10
1 ≤ Sum of timei ≤ 1000000
Sample Input 1
5 100
p1 150
p2 80
p3 200
p4 350
p5 20
Sample Output 1
p2 180
p5 400
p1 450
p3 550
p4 800
Notes
Template in C
code
/*
^....0
^ .1 ^1^
.. 01
1.^ 1.0
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.1 ^0 .........001^
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00 ^ 11^ ^1. .1^
1.^ ^0 0^
.^ ^0..1
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1 .0 ^ ^
^ 00. ^^0.^
1 ^ 0 ^^110.^
0. 0 ^ ^^^10.01
^^ 010^ 1 1 ^^^1110.1
0001 10 0 ^ 1.1 ^^^1111110
0^ 10 . 01 ^^ ^^ ^^^1111^1.^ ^^^
10 10^ 0^ ^^111^^^0.1^ 1....^
11 0 ^^11^^^ 0.. ....1^ ^ ^
1. 0^ ^11^^^ ^ 1 111^ ^ 0.
10 00 11 ^^^^^ 1 0 1.
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0^ 1.0 .^ ^^^^ 1 1 .0
^.^ ^^ 0^ ^1 ^^^^ 0. ^.1
1 ^ 11 1. ^^^ ^ ^ ..^
^..^ ^1 ^.^ ^^^ .0 ^.0
0..^ ^0 01 ^^^ .. 0..^
1 .. .1 ^.^ ^^^ 1 ^ ^0001
^ 1. 00 0. ^^^ ^.0 ^.1
. 0^. ^.^ ^.^ ^^^ ..0.0
1 .^^. .^ 1001 ^^ ^^^ . 1^
. ^ ^. 11 0. 1 ^ ^^ 0.
0 ^. 0 ^0 1 ^^^ 0.
0.^ 1. 0^ 0 .1 ^^^ ..
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.0 00 . ^^^^^^ .
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^.1 00 ^. ^^^ .1
.. 00 .1 1..01 ..
1.1 00 1. ..^ 10
^ 1^ 00 ^.1 0 1 1
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. 00 ^.^ 10^ ^^
1.1 00 00 10^
..^ 1. ^. 1.
0 1 ^. 00 00 .^
^ ^. ^ 1 00 ^0000^ ^ 01
1 0 ^. 00.0^ ^00000 1.00.1 11
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1 1 0.
.. 1 ^
1 1
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^ 0 1..01^^100000..0^
1 1 ^ 1 ^^1111^ ^^
0 ^ ^ 1 1000^
.1 ^.^ . 00
.. 1.1 0. 0
1. . 1. .^
1. 1 1. ^0
^ . ^.1 00 01
^.0 001. .^
*/
// Virtual_Judge —— Queue Aizu - ALDS1_3_B.cpp created by VB_KoKing on 2019,04,29,07.
/* Procedural objectives：
Variables required by the program:
Procedural thinking：
Functions required by the program:
*/
/* My dear Max said：
"I like you,
So the first bunch of sunshine I saw in the morning is you,
The first hurricane that passed through your ear is you,
The first star you see is also you.
The world I see is all your shadow."
FIGHTING FOR OUR FUTURE！！！
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#define LEN 100007
using namespace std;
//代表任务的结构体
typedef struct pp
{
char name[100];
int t;
}P;
P Q[LEN];
int head,tail,n;
void enqueue(P x)
{
Q[tail]=x;
tail = (tail+1) % LEN;
}
P dequeue()
{
P x=Q[head];
head=(head+1)%LEN;
return x;
}
int main()
{
int elaps=0,q;
P u;
cin>>n>>q;
//按顺序将所有任务添加至队列
for (int i = 1; i <= n; i++)
cin>>Q[i].name>>Q[i].t;
head=1;tail=n+1;
//模拟
while (head!=tail)
{
u=dequeue();
int c=min(q,u.t);
u.t-=c;elaps+=c;
if (u.t>0) enqueue(u);
else cout<<u.name<<' '<<elaps<<endl;
}
return 0;
}
