PAT1017:Queueing at Bank
1017. Queueing at Bank (25)
Suppose a bank has K windows open for service. There is a yellow line in front of the windows which devides the waiting area into two parts. All the customers have to wait in line behind the yellow line, until it is his/her turn to be served and there is a window available. It is assumed that no window can be occupied by a single customer for more than 1 hour.
Now given the arriving time T and the processing time P of each customer, you are supposed to tell the average waiting time of all the customers.
Input Specification:
Each input file contains one test case. For each case, the first line contains 2 numbers: N (<=10000) - the total number of customers, and K (<=100) - the number of windows. Then N lines follow, each contains 2 times: HH:MM:SS - the arriving time, and P - the processing time in minutes of a customer. Here HH is in the range [00, 23], MM and SS are both in [00, 59]. It is assumed that no two customers arrives at the same time.
Notice that the bank opens from 08:00 to 17:00. Anyone arrives early will have to wait in line till 08:00, and anyone comes too late (at or after 17:00:01) will not be served nor counted into the average.
Output Specification:
For each test case, print in one line the average waiting time of all the customers, in minutes and accurate up to 1 decimal place.
Sample Input:
7 3
07:55:00 16
17:00:01 2
07:59:59 15
08:01:00 60
08:00:00 30
08:00:02 2
08:03:00 10
Sample Output:
8.2 思路
贪心策略,将办理业务的人按到达时间递增排序,先来的优先服务(将时间统一转换为秒方便计算)。
注意:
1.业务办理时间不应超过1小时。
2.17点之后来的人不提供服务。
3.一个窗口的空闲时刻有两种情况:
1)空闲直到下一个人customer[i]到达银行办理业务,那么这个窗口的下一个空闲时刻为customer[i]的到达时间加上他办理业务需要的时间。
2)如果在一个窗口空闲前,custmoer[i]就先来了,那么他需要等(窗口空闲时间-他到达的时间)这么一段时间。该窗口的下一个空闲时刻就是它当前空闲时刻加上customer[i]的业务办理时间。 代码
#include<iostream>
#include<vector>
#include<algorithm>
#include<iomanip>
using namespace std;
class person
{
public:
int cometime;
int taketime;
person(int ct,int tt)
{
cometime = ct;
taketime = tt;
}
}; /*贪心策略 先来的先服务,有空窗口就服务*/ bool cmp(const person& a,const person& b)
{
return a.cometime < b.cometime;
} int main()
{
int N,K;
vector<person> customer;
while(cin >> N >> K)
{
vector<int> windows(K,28800);
for(int i = 0;i < N;i++)
{
int hh,mm,ss,lasttime;
scanf("%d:%d:%d %d",&hh,&mm,&ss,&lasttime);
if(lasttime > 60)
lasttime = 60;
lasttime *= 60;
int arrivetime = hh * 3600 + mm * 60 + ss;
if(arrivetime > 61200)
continue;
customer.push_back(person(arrivetime,lasttime));
}
sort(customer.begin(),customer.end(),cmp);
double waitTime = 0;
for(int i = 0;i < customer.size();i++)
{
int minwindow = windows[0],tmpindex = 0;
for(int j = 0;j < K;j++)
{
if(windows[j] < minwindow)
{
minwindow = windows[j];
tmpindex = j;
}
}
if(customer[i].cometime >= minwindow)
{
windows[tmpindex] = customer[i].cometime + customer[i].taketime;
}
else
{
waitTime += minwindow - customer[i].cometime;
windows[tmpindex] += customer[i].taketime;
}
}
if(customer.empty())
cout << "0.0" << endl;
else
cout << fixed << setprecision(1) << waitTime/60.0/customer.size() << endl;
}
}
PAT1017:Queueing at Bank的更多相关文章
- pat1017. Queueing at Bank (25)
1017. Queueing at Bank (25) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Supp ...
- PAT甲级1017. Queueing at Bank
PAT甲级1017. Queueing at Bank 题意: 假设一家银行有K台开放服务.窗前有一条黄线,将等候区分为两部分.所有的客户都必须在黄线后面排队,直到他/她轮到服务,并有一个可用的窗口. ...
- PAT 1017 Queueing at Bank[一般]
1017 Queueing at Bank (25)(25 分)提问 Suppose a bank has K windows open for service. There is a yellow ...
- PAT 1017 Queueing at Bank (模拟)
1017. Queueing at Bank (25) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Supp ...
- PAT 甲级 1017 Queueing at Bank (25 分)(模拟题,有点思维小技巧,第二次做才理清思路)
1017 Queueing at Bank (25 分) Suppose a bank has K windows open for service. There is a yellow line ...
- pat——1017. Queueing at Bank (java中Map用法)
由PAT1017例题展开: Suppose a bank has K windows open for service. There is a yellow line in front of the ...
- 1017. Queueing at Bank (25)
Suppose a bank has K windows open for service. There is a yellow line in front of the windows which ...
- PAT 1017. Queueing at Bank
Suppose a bank has K windows open for service. There is a yellow line in front of the windows which ...
- 1017. Queueing at Bank (25) - priority_queuet
题目如下: Suppose a bank has K windows open for service. There is a yellow line in front of the windows ...
随机推荐
- 理解WebKit和Chromium: 网页渲染的基本过程
转载请注明原文地址:http://blog.csdn.net/milado_nju ## 概述 前面介绍了一些渲染引擎的功能,包括网络,资源加载,DOM树,RenderObject树等等,但是,给人以 ...
- Ubuntu下编译SHTOOLS
SHTOOLS是使用Fortran语言写的一个专门用于处理球谐函数的一个开源库,更多的介绍请猛戳这里,关于这个库的安装和使用,都在官网上有详细的说明,虽然很详细,但是编译的时候还是比较费劲,下面将我在 ...
- Xcode中的调试工具栏简介
如下图所示: 从左至右,第一个按钮用来隐藏调试区域. 第二个按钮向你展示断点是否被全局开启或禁用.如果它不是高亮蓝色,则没有断点会被触发. 第三个按钮暂停或继续程序的执行,你一般点击它继续运行到程序的 ...
- 【Qt编程】Qt学习之Window and Dialog Widgets
Qt Creator 提供的默认基类只要QMainWindow.QWidget和QDialog三种.其中,QMainWindow是带有菜单栏和工具栏的主窗口类,QDialog是各种对话框的基类,这两个 ...
- 第十一章 图像之2D(1)SpriteBatch
Android游戏开发群:290051794 Libgdx游戏开发框架交流群:261954621 作者:宋志辉 出处:http://blog.csdn.net/song19891121 本文版权归作 ...
- Android高效率编码-第三方SDK详解系列(一)——百度地图,绘制,覆盖物,导航,定位,细腻分解!
Android高效率编码-第三方SDK详解系列(一)--百度地图,绘制,覆盖物,导航,定位,细腻分解! 这是一个系列,但是我也不确定具体会更新多少期,最近很忙,主要还是效率的问题,所以一些有效的东西还 ...
- Windows核心编程读书笔记1
今天特别困啊,这是为什么?!!刚刚把第一章看了一下,困到不行,所以写blog清醒一下. 第一章标题是“错误处理”,看了之后吓了一跳,难道第一章就讲这么高大上的东西?!不是不是,我现在的理解是,这章主要 ...
- (WPS) 网络地理信息处理服务
WPS 标准为网络地理信息处理服务提供了标准化的输入和输出. OGC® Web Processing Service (WPS) 标准描述了如何通过远程的任何算法和模型处理获得地理空间的栅格或矢量信息 ...
- sqlplus 登录数据库
sqlplus pams/pamscncc@ORCLMIS
- struts2线程安全
struts2线程安全 2012-02-16 21:07:58 分类: 系统运维 问题:Struts 2 Action对象为每一个请求产生一个实例,因此没有线程安全问题.Spring的Ioc容器管理 ...