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

  1. 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 ...

  2. 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 ...

  3. 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 ...

  4. 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 ...

  5. 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 ...

  6. 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. ...

  7. 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 ...

  8. PAT 甲级 1017 Queueing at Bank

    https://pintia.cn/problem-sets/994805342720868352/problems/994805491530579968 Suppose a bank has K w ...

  9. 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

  1. PHP PhantomJs中文文档(翻译)

    介绍 PHP PhantomJS 是一个灵活的 PHP 库加载页面通过 PhantomJS 无头浏览器并将返回页面响应.这是方便于需要JavaScript的支持,同时还支持截屏测试网站.功能列表通过 ...

  2. Python 命令行参数解析

    方法1: Python有一个类可以专门处理命令行参数,先看代码: #!/usr/bin/env python # encoding: utf-8 from optparse import Option ...

  3. 对ARM9哈佛结构的认识

    书本上都说ARM是哈佛结构,但是我总感觉好像看不出来.后来针对S3C2440的ARM9核进行分析,我有了自己的见解. 我的结论是“ARM9被称为是哈佛结构是从它拥有指令cache和数据cache”来说 ...

  4. HTML -- 元素和属性

    HTML -- 元素 HTML元素是从开始标签到结束标签之间的代码,如: <!-- 加粗标签 --> <b>一些元素</b> <!-- 换行 --> & ...

  5. cocos2dx 3.3创建新项目 和 VS2012解决方案加载失败问题

     首先创建新项目,步骤如下: 1.进入cocos2d-x-3.3\tools\cocos2d-console\bin目录,按住shift+鼠标右键 2.输入 cocos new 项目名 –p 包名 – ...

  6. Xcode 证书生成、设置、应用

    Xcode 证书生成.设置.应用,与大家分享.如果按下面步骤还不能编译成功,我手把手教你. 开发环境: Mac OS lion 10.7.4 XCode 4.3.3 1         点击钥匙图标 ...

  7. SQL Server强制删除复制发布

    原文地址:http://blog.csdn.net/leamonjxl/article/details/7352208 SQL Server 中 存在以前(系统还原前)的发布内容,使用鼠标->右 ...

  8. 在C#里实现各种窗口切换特效,多达13种特效

    原文:http://www.cnblogs.com/clayui/archive/2011/06/28/2092126.html 预览:   下载 这次clayui给大家带来了比较实用的东西,因为时间 ...

  9. Android 使用XmlSerializer生成xml文件

    在Android开发中,我们时常要用到xml文件. xml作为一种数据载体,在数据传输中发挥着重要的作用,而且它可读性比较强. 下面给出在Android开发中使用XmlSerializer类生成一个简 ...

  10. Jquery UI dialog 传参

    [一篮饭特稀原创,转载请注明出自http://www.cnblogs.com/wanghafan/p/3519318.html] $("#dialog").dialog({ aut ...