pat1014. Waiting in Line (30)
1014. Waiting in Line (30)
Suppose a bank has N windows open for service. There is a yellow line in front of the windows which devides the waiting area into two parts. The rules for the customers to wait in line are:
- The space inside the yellow line in front of each window is enough to contain a line with M customers. Hence when all the N lines are full, all the customers after (and including) the (NM+1)st one will have to wait in a line behind the yellow line.
- Each customer will choose the shortest line to wait in when crossing the yellow line. If there are two or more lines with the same length, the customer will always choose the window with the smallest number.
- Customer[i] will take T[i] minutes to have his/her transaction processed.
- The first N customers are assumed to be served at 8:00am.
Now given the processing time of each customer, you are supposed to tell the exact time at which a customer has his/her business done.
For example, suppose that a bank has 2 windows and each window may have 2 custmers waiting inside the yellow line. There are 5 customers waiting with transactions taking 1, 2, 6, 4 and 3 minutes, respectively. At 08:00 in the morning, customer1 is served at window1 while customer2 is served at window2. Customer3 will wait in front of window1 and customer4 will wait in front of window2. Customer5 will wait behind the yellow line.
At 08:01, customer1 is done and customer5 enters the line in front of window1 since that line seems shorter now. Customer2 will leave at 08:02, customer4 at 08:06, customer3 at 08:07, and finally customer5 at 08:10.
Input
Each input file contains one test case. Each case starts with a line containing 4 positive integers: N (<=20, number of windows), M (<=10, the maximum capacity of each line inside the yellow line), K (<=1000, number of customers), and Q (<=1000, number of customer queries).
The next line contains K positive integers, which are the processing time of the K customers.
The last line contains Q positive integers, which represent the customers who are asking about the time they can have their transactions done. The customers are numbered from 1 to K.
Output
For each of the Q customers, print in one line the time at which his/her transaction is finished, in the format HH:MM where HH is in [08, 17] and MM is in [00, 59]. Note that since the bank is closed everyday after 17:00, for those customers who cannot be served before 17:00, you must output "Sorry" instead.
Sample Input
2 2 7 5
1 2 6 4 3 534 2
3 4 5 6 7
Sample Output
08:07
08:06
08:10
17:00
Sorry
教训:
Note that since the bank is closed everyday after 17:00, for those customers who cannot be served before 17:00, you must output "Sorry" instead.
这句话的含义是如果不能在17:00之前开始服务,则输出“Sorry”
注意:vector就是可变长的动态数组,比较灵活好用
加入元素:push_back()
删除元素:用vector<int>::iterator it 遍历至要删除的元素,然后v.erase(it)
元素个数:size()
访问元素:和一般的数组一样,直接访问
#include <cstdio>
#include <cstring>
#include <string>
#include <queue>
#include <vector>
#include <iostream>
using namespace std;
struct custom{
int cost,finish;
};
vector<int> v[];
custom cu[];
int main(){
//freopen("D:\\INPUT.txt","r",stdin);
int n,m,k,q;
scanf("%d %d %d %d",&n,&m,&k,&q);
int i,j;
for(i=;i<k;i++){
scanf("%d",&cu[i].cost);
}
for(i=;i<n&&i<k;i++){
cu[i].finish=cu[i].cost;
v[i].push_back(i);
}
for(;i<m*n&&i<k;i++){
cu[i].finish=cu[v[i%n][v[i%n].size()-]].finish+cu[i].cost;
v[i%n].push_back(i);
}
for(;i<k;i++){
int mintime=cu[v[][]].finish,minnum=;
for(j=;j<n;j++){
if(cu[v[j][]].finish<mintime){
minnum=j;
mintime=cu[v[j][]].finish;
}
}
cu[i].finish=cu[v[minnum][v[minnum].size()-]].finish+cu[i].cost;
vector<int>::iterator it=v[minnum].begin();
v[minnum].erase(it);
v[minnum].push_back(i);
}
int num;
for(i=;i<q;i++){
scanf("%d",&num);
if(cu[num-].finish-cu[num-].cost<){
//cout<<num-1<<" "<<cu[num-1].finish<<endl;
int h=cu[num-].finish/+;
if(h>){
continue;
}
int m=cu[num-].finish%;
if(h>){
cout<<h;
}
else{
cout<<<<h;
}
cout<<":";
if(m>){
cout<<m;
}
else{
cout<<<<m;
}
cout<<endl;
}
else{
cout<<"Sorry"<<endl;
}
}
return ;
}
pat1014. Waiting in Line (30)的更多相关文章
- PAT-1014 Waiting in Line (30 分) 优先队列
Suppose a bank has N windows open for service. There is a yellow line in front of the windows which ...
- PAT 甲级 1014 Waiting in Line (30 分)(queue的使用,模拟题,有个大坑)
1014 Waiting in Line (30 分) Suppose a bank has N windows open for service. There is a yellow line ...
- 1014 Waiting in Line (30分)
1014 Waiting in Line (30分) Suppose a bank has N windows open for service. There is a yellow line i ...
- 1014. Waiting in Line (30)
Suppose a bank has N windows open for service. There is a yellow line in front of the windows which ...
- 1014 Waiting in Line (30)(30 point(s))
problem Suppose a bank has N windows open for service. There is a yellow line in front of the window ...
- 1014 Waiting in Line (30)(30 分)
Suppose a bank has N windows open for service. There is a yellow line in front of the windows which ...
- 1014 Waiting in Line (30 分)
Suppose a bank has N windows open for service. There is a yellow line in front of the windows which ...
- PTA 1014 Waiting in Line (30分) 解题思路及满分代码
题目 Suppose a bank has N windows open for service. There is a yellow line in front of the windows whi ...
- PAT A 1014. Waiting in Line (30)【队列模拟】
题目:https://www.patest.cn/contests/pat-a-practise/1014 思路: 直接模拟类的题. 线内的各个窗口各为一个队,线外的为一个,按时间模拟出队.入队. 注 ...
随机推荐
- 函数返回值string与返回值bool区别------c++程序设计原理与实践(进阶篇)
为什么find_from_addr()和find_subject()如此不同?比如,find_from_addr()返回bool值,而find_subject()返回string.原因在于我们想说明: ...
- iOS App 内部跳转(设置、Wifi、蓝牙...)关键词
1.iOS 10 以前: 蜂窝网络:prefs:root=MOBILE_DATA_SETTINGS_ID Wi-Fi:prefs:root=WIFI 定位服务:prefs:root=LOCATION_ ...
- 可变大小、颜色边框、样式的UISwitch
1.CHSwitch.h // // 文 件 名:CHSwitch.h // // 版权所有:Copyright © 2018 lelight. All rights reserved. // 创 建 ...
- 详说Flask、Django、Pyramid三大主流 Web 框架
前言 目前随着 Python 在大数据.云计算.人工智能方面的热度,Python Web 应该也会被更多企业了解使用. Python Web 框架千万种,没必要都去了解和学习,身边总有人说高手都用 F ...
- postgreSQL PL/SQL编程学习笔记(六)——杂七杂八
1 PL/pgSQL Under the Hood This part discusses some implementation details that are frequently import ...
- BUAA_OO_电梯系列
电梯作业 第一次作业和第二次作业 由于我第一次作业给傻瓜电梯写了捎带所以第一次第二次作业差不多 电梯运行一个线程Elevator,输入控制一个线程Call 一个物理电梯控制表可以完成移动和进出人功能, ...
- SDUT OJ 学密码学一定得学程序
学密码学一定得学程序 Time Limit: 1000 ms Memory Limit: 65536 KiB Submit Statistic Discuss Problem Description ...
- 蓝牙4.0BLE抓包(三) – 扫描请求和扫描响应
版权声明:本文为博主原创文章,转载请注明作者和出处. 作者:强光手电[艾克姆科技-无线事业部] 1. 扫描请求和扫描响应 广播包含扫描请求SCAN_REQ和扫描响应SCAN_RSP. 扫描请求: ...
- Android 单选按钮(RadioButton)和复选框(CheckBox)的使用
1.RadioButton (1)介绍 (2)单选按钮点击事件的用法 (3)RadioButton与RadioGroup配合使用实现单选题功能 (4)xml布局及使用 <?xml version ...
- Android 应用资源及R文件的位置
1.介绍 (1)常识 (2)在res目录下新建资源文件(例如数字资源) app--->res,选择res,右击new--->value resource file 2.字符资源(strin ...