当前位置: 首页 > news >正文

医院设备利用(Use of Hospital Facilities, ACM/ICPC World Finals 1991, UVa212)rust解法

医院里有n(n≤10)个手术室和m(m≤30)个恢复室。每个病人首先会被分配到一个手术室,手术后会被分配到一个恢复室。从任意手术室到任意恢复室的时间均为t1,准备一个手术室和恢复室的时间分别为t2和t3(一开始所有手术室和恢复室均准备好,只有接待完一个病人之后才需要为下一个病人准备)。
k名(k≤100)病人按照花名册顺序排队,T点钟准时开放手术室。每当有准备好的手术室时,队首病人进入其中编号最小的手术室。手术结束后,病人应立刻进入编号最小的恢复室。如果有多个病人同时结束手术,在编号较小的手术室做手术的病人优先进入编号较小的恢复室。输入保证病人无须排队等待恢复室。
输入n、m、T、t1、t2、t3、k和k名病人的名字、手术时间和恢复时间,模拟这个过程。

样例:
输入

5 12 07 5 15 10 16
Jones
28 140
Smith
120 200
Thompson
23 75
Albright
19 82
Poucher
133 209
Comer
74 101
Perry
93 188
Page
111 223
Roggio
69 122
Brigham
42 79
Nute
22 71
Young
38 140
Bush
26 121
Cates
120 248
Johnson
86 181
White
92 140

输出

 Patient          Operating Room          Recovery Room#  Name     Room#  Begin   End      Bed#  Begin    End------------------------------------------------------1  Jones      1    7:00    7:28      3    7:33    9:532  Smith      2    7:00    9:00      1    9:05   12:253  Thompson   3    7:00    7:23      2    7:28    8:434  Albright   4    7:00    7:19      1    7:24    8:465  Poucher    5    7:00    9:13      5    9:18   12:476  Comer      4    7:34    8:48      4    8:53   10:347  Perry      3    7:38    9:11      2    9:16   12:248  Page       1    7:43    9:34      6    9:39   13:229  Roggio     4    9:03   10:12      9   10:17   12:19
10  Brigham    2    9:15    9:57      8   10:02   11:21
11  Nute       3    9:26    9:48      7    9:53   11:04
12  Young      5    9:28   10:06      3   10:11   12:31
13  Bush       1    9:49   10:15     10   10:20   12:21
14  Cates      3   10:03   12:03      8   12:08   16:16
15  Johnson    2   10:12   11:38      4   11:43   14:44
16  White      5   10:21   11:53      7   11:58   14:18Facility Utilization
Type  # Minutes  % Used
-------------------------
Room  1     165   29.68
Room  2     248   44.60
Room  3     258   46.40
Room  4     162   29.14
Room  5     263   47.30
Bed   1     282   50.72
Bed   2     263   47.30
Bed   3     280   50.36
Bed   4     282   50.72
Bed   5     209   37.59
Bed   6     223   40.11
Bed   7     211   37.95
Bed   8     327   58.81
Bed   9     122   21.94
Bed  10     121   21.76
Bed  11       0    0.00
Bed  12       0    0.00

解法:

use std::{cmp::Ordering, collections::BinaryHeap, io};
#[derive(Debug)]
struct Patient {id: usize,name: String,operation_time: usize,recovery_time: usize,room_id: usize,room_begin_time: usize,room_end_time: usize,bed_id: usize,bed_begin_time: usize,bed_end_time: usize,
}
#[derive(Debug, PartialEq, Eq)]
struct Room {id: usize,start_available_time: usize,
}
impl Ord for Room {fn cmp(&self, other: &Self) -> std::cmp::Ordering {if self.start_available_time != other.start_available_time {if self.start_available_time < other.start_available_time {Ordering::Greater} else {Ordering::Less}} else if self.id < other.id {Ordering::Greater} else {Ordering::Less}}
}
impl PartialOrd for Room {fn partial_cmp(&self, other: &Self) -> Option<Ordering> {Some(self.cmp(other))}
}fn main() {let mut buf = String::new();io::stdin().read_line(&mut buf).unwrap();let v: Vec<usize> = buf.split_whitespace().map(|e| e.parse().unwrap()).collect();let n = v[0];let m = v[1];let start_time = v[2] * 60;let t1 = v[3];let t2 = v[4];let t3 = v[5];let k = v[6];let mut rooms = BinaryHeap::new();for i in 0..n {let room = Room {id: i + 1,start_available_time: start_time,};rooms.push(room);}let mut patients: Vec<Patient> = vec![];for i in 0..k {let mut buf = String::new();io::stdin().read_line(&mut buf).unwrap();let name = buf.trim().to_string();let mut buf = String::new();io::stdin().read_line(&mut buf).unwrap();let t: Vec<usize> = buf.split_whitespace().map(|e| e.parse().unwrap()).collect();let patient = Patient {id: i + 1,name: name,operation_time: t[0],recovery_time: t[1],room_id: 0,room_begin_time: 0,room_end_time: 0,bed_id: 0,bed_begin_time: 0,bed_end_time: 0,};patients.push(patient);}let mut room_used_time: Vec<usize> = vec![0; n];for patient in patients.iter_mut() {let aroom = rooms.pop().unwrap();patient.room_id = aroom.id;patient.room_begin_time = aroom.start_available_time;patient.room_end_time = patient.room_begin_time + patient.operation_time;rooms.push(Room {id: aroom.id,start_available_time: patient.room_end_time + t2,});room_used_time[aroom.id - 1] += patient.operation_time;}patients.sort_by(|a, b| {if a.room_end_time != b.room_end_time {if a.room_end_time < b.room_end_time {Ordering::Less} else {Ordering::Greater}} else {if a.room_id < b.room_id {Ordering::Less} else {Ordering::Greater}}});let mut beds: Vec<usize> = vec![start_time; m];let mut bed_used_time: Vec<usize> = vec![0; m];let mut final_time = start_time;for patient in patients.iter_mut() {for j in 0..m {if beds[j] <= patient.room_end_time {patient.bed_id = j + 1;patient.bed_begin_time = patient.room_end_time + t1;patient.bed_end_time = patient.bed_begin_time + patient.recovery_time;beds[j] = patient.bed_end_time + t3;bed_used_time[j] += patient.recovery_time;final_time = final_time.max(patient.bed_end_time);break;}}}patients.sort_by(|a, b| a.id.cmp(&b.id));println!(" Patient          Operating Room          Recovery Room");println!(" #  Name     Room#  Begin   End      Bed#  Begin    End");println!(" ------------------------------------------------------");for i in patients.iter() {print!("{:2}  {:<10}", i.id, i.name);print!("{:2}   {:2}:{:02}   {:2}:{:02}",i.room_id,i.room_begin_time / 60,i.room_begin_time % 60,i.room_end_time / 60,i.room_end_time % 60);println!("     {:2}   {:2}:{:02}   {:2}:{:02}",i.bed_id,i.bed_begin_time / 60,i.bed_begin_time % 60,i.bed_end_time / 60,i.bed_end_time % 60);}println!();println!("Facility Utilization");println!("Type  # Minutes  % Used");println!("-------------------------");let all_time = final_time - start_time;for i in 0..n {println!("Room {:2}    {:4}   {:5.2}",i + 1,room_used_time[i],100.0 * room_used_time[i] as f64 / all_time as f64);}for i in 0..m {println!("Bed  {:2}    {:4}   {:5.2}",i + 1,bed_used_time[i],100.0 * bed_used_time[i] as f64 / all_time as f64);}
}
http://www.lryc.cn/news/204137.html

相关文章:

  • 解决github ping不通的问题(1024程序员节快乐!
  • QT基础 柱状图
  • 微机原理与接口技术-第七章输入输出接口
  • YoloV8改进策略:独家原创,LSKA(大可分离核注意力)改进YoloV8,比Transformer更有效,包括论文翻译和实验结果
  • 7天易语言从入门到实战(一)
  • redis缓存问题
  • mysql创建自定义函数报错
  • Docker 的数据管理与网络通信以及Docker镜像的创建
  • linux系统查看bash的history
  • 【T+】畅捷通T+增加会计科目提示执行超时已过期。
  • 0基础学习VR全景平台篇第111篇:全景图拼接和编辑 - PTGui Pro教程
  • Dynamics 365 使用ILMerge 合并CRM开发后的DLL
  • SpringBoot Web请求响应
  • Jenkins CLI二次开发工具类
  • 2. 计算WPL
  • 筹备三年,自动驾驶L3标准将至,智驾产业链的关键一跃
  • 【Python】Python使用Switch语句
  • 一年一度的1024程序员节
  • 第十七章 数据库操作
  • RTI-DDS代码分析使用介绍
  • ant-design-vue 3 a-table保留选中状态
  • golang 工程组件:grpc-gateway option自定义http规则
  • 亚马逊添加购物车和收藏有什么区别
  • JAVA-编程基础-11-03-java IO 字节流
  • python之Cp、Cpk、Pp、Ppk
  • 统信uos 1030 企业版 安装.net core环境
  • 2023/10/23学习记录
  • flask入门(四)前后端数据传输
  • JS——垃圾回收的原理
  • Spring Cloud Gateway 路由构建器的源码分析