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 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
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std; int N,K,wait_time=;
struct Customer
{
int arrive_time;
int need_time;
}; struct Customer customer[]; struct Windows
{
int next_available_time;
}; struct Windows windows[]; bool cmp(struct Customer a,struct Customer b)
{
return a.arrive_time<b.arrive_time;
} int find_available_windows(int arrive_time)
{
int i;
for(i=;i<K;i++) {
if(windows[i].next_available_time<=arrive_time) {
return i;
}
}
return -;
} int find_earliest_window()
{
int i;
int e=;
for(i=;i<K;i++) {
if(windows[i].next_available_time<windows[e].next_available_time) {
e=i;
}
}
return e;
} int main()
{
scanf("%d%d",&N,&K);
int i;
char arrive_time[];
int need_time;
for(i=;i<K;i++)
windows[i].next_available_time=*;
int len=;
for(i=;i<N;i++) {
int h,m,s;
scanf("%s%d",arrive_time,&need_time);
if(strcmp(arrive_time,"17:00:00")>)
continue; sscanf(arrive_time,"%d:%d:%d",&h,&m,&s);
if(h<)
wait_time+=*-(*h+*m+s);
customer[len].arrive_time=*h+*m+s;
customer[len++].need_time=need_time*;
}
N=len; sort(customer,customer+N,cmp); for(i=;i<N;i++) {
int w=find_available_windows(customer[i].arrive_time);
if(w>=) {//找到空闲窗口
// windows[w].next_available_time=customer[i].arrive_time+customer[i].need_time;
if(customer[i].arrive_time<*) {
windows[w].next_available_time=*+customer[i].need_time;
} else {
windows[w].next_available_time=customer[i].arrive_time+customer[i].need_time;
}
} else { //找不到空闲窗口
w=find_earliest_window();
/* wait_time+=windows[w].next_available_time-customer[i].arrive_time;
* windows[w].next_available_time=(windows[w].next_available_time-customer[i].arrive_time)+customer[i].need_time;
*/
if(customer[i].arrive_time<*) {//如果到得早 窗口的下个可用时间等于当前下个可用时间加新来顾客所需要服务时间
wait_time+=windows[w].next_available_time-*;
windows[w].next_available_time=windows[w].next_available_time+customer[i].need_time;
} else {
wait_time+=windows[w].next_available_time-customer[i].arrive_time;
windows[w].next_available_time=windows[w].next_available_time+customer[i].need_time;
} }
} printf("%.1f\n",1.0*wait_time/60.0/N);
}
PAT 1017. Queueing at Bank的更多相关文章
- 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——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 ...
- PAT 1017 Queueing at Bank (25) (坑题)
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 (模拟)
Suppose a bank has K windows open for service. There is a yellow line in front of the windows which ...
- 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 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 甲级 1017 Queueing at Bank
https://pintia.cn/problem-sets/994805342720868352/problems/994805491530579968 Suppose a bank has K w ...
- PAT (Advanced Level) 1017. Queueing at Bank (25)
简单模拟. #include<iostream> #include<cstring> #include<cmath> #include<algorithm&g ...
随机推荐
- ASP.NET 共用类库
using System; using System.Collections.Generic; using System.Text; using System.Web; using System.We ...
- Excel Skill (1) -- 判断时如何去掉框里的空格
使用命令 TRIM 说明: Purpose. Remove extra spaces from text. Text with extra spaces removed. =TRIM (text) t ...
- 国内外最全的asp.net开源项目 (转)
最近一些项目开始用到CMS系统,最开始是研究JAVA的,无奈国内JAVA的CMS开源系统还是比较少,最多最成熟的还是PHP的,当然现在.NET的也不少了,这里做一下汇总备忘,留待学习研究. 国内系统: ...
- 时序图(Sequence Diagram)
控制焦点Focus on Control 的取值: Alternative fragment(denoted “alt”) 与 if…then…else对应 Option fragment (deno ...
- ImageButton和Button区别
一.基础准备 Imagebutton 继承 Imageview,就是用一个图标代表了一些文字,它没Android:text属性.它由Android:src指定图标的位置 android:s ...
- js node
http://blogs.msdn.com/b/scott_hanselman/archive/2011/11/29/window-iis-node-js.aspx http://www.16kan. ...
- 【UVA 10369】 Arctic Network (最小生成树)
[题意] 南极有n个科研站, 要把这些站用卫星或者无线电连接起来,使得任意两个都能直接或者间接相连.任意两个都有安装卫星设备的,都可以直接通过卫星通信,不管它们距离有多远. 而安装有无线电设备的两个站 ...
- QWidget与HWND的互相转换
QWidget与HWND的互相转换 在编写Windows的应用程序时,我们有时不可避免地要与Windows平台固有的Win32 API打交道,但是Win32 API里面常常用到的HWND等诸多句柄QT ...
- Hadoop开发环境搭建
hadoop是一个分布式系统基础架构,由Apache基金会所开发. 用户可以在不了解分布式底层细节的情况下,开发分布式程序.充分利用集群的威力高速运算和存储. Hadoop实现了一个分布式文件系统 ...
- android使用tabhost实现导航
参考 http://blog.csdn.net/xixinyan/article/details/6771341 http://blog.sina.com.cn/s/blog_6b04c8eb0101 ...