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. Homebrew安装php5及composer for mac教程

    安装brew 可以查看教程:mac os x 10.9.1 安装 Homebrew软件包管理工具及brew安装maven3.1.1 首先更新下brew软件库 brew update brew tap ...

  2. Remark of BLENDFUNCTION from MSDN

    Remarks When the AlphaFormat member is AC_SRC_ALPHA, the source bitmap must be 32 bpp. If it is not, ...

  3. 预处理命令#define #undef #if #endif 的基本用法

    C#的预处理命令其实还是蛮有用的,但是真正使用过得人不多,这个介绍一下平时用的比较多的预处理命令中的几个:#define,#undef ,#if,#endif.除此之外还有一些预处理命令#warnin ...

  4. python中的__init__ 、__new__、__call__等内置函数的剖析

    1.__new__(cls, *args, **kwargs)   创建对象时调用,返回当前对象的一个实例;注意:这里的第一个参数是cls即class本身2.__init__(self, *args, ...

  5. PYTHON之DEF

    def sayHello(): print('Hello World!') while True: s = input('Enter something : ') if s == 'quit': br ...

  6. 【网络流24题】No.18 分配问题 (二分图最佳匹配 费用流|KM)

    [题意] 有 n 件工作要分配给 n 个人做.第 i 个人做第 j 件工作产生的效益为 cij . 试设计一个将n 件工作分配给 n 个人做的分配方案, 使产生的总效益最大. 输入文件示例input. ...

  7. 【网络流24题】 No.15 汽车加油行驶问题 (分层图最短路i)

    [题意] 问题描述:给定一个 N*N 的方形网格,设其左上角为起点◎, 坐标为( 1, 1), X 轴向右为正, Y轴向下为正, 每个方格边长为 1, 如图所示. 一辆汽车从起点◎出发驶向右下角终点▲ ...

  8. [转贴] C/C++中动态链接库的创建和调用

    DLL 有助于共享数据和资源.多个应用程序可同时访问内存中单个DLL 副本的内容.DLL 是一个包含可由多个程序同时使用的代码和数据的库.下面为你介绍C/C++中动态链接库的创建和调用. 动态连接库的 ...

  9. SPRING IN ACTION 第4版笔记-第二章-004-Bean是否单例

    spring的bean默认是单例,加载容器是会被化,spring会拦截其他再次请求bean的操作,返回spring已经创建好的bean. It appears that the CompactDisc ...

  10. 远程仓库版本回退方法 good

    1 简介 最近在使用git时遇到了远程分支需要版本回滚的情况,于是做了一下研究,写下这篇博客. 2 问题 如果提交了一个错误的版本,怎么回退版本? 如果提交了一个错误的版本到远程分支,怎么回退远程分支 ...