PAT甲级练习1014. Waiting in Line (30)
来源:互联网 发布:看舌头知病情图片 编辑:程序博客网 时间:2024/06/06 07:53
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 Input2 2 7 51 2 6 4 3 534 23 4 5 6 7Sample Output
08:0708:0608:1017:00Sorry
主要思路:对于n*m的人来说直接按顺序排上去就好。之后的人要根据各个窗口先出现空先排。对每个窗口使用一个queue来保存排队时间,startTime记录新来的人被开始服务的时间,popTime记录队首的人离开的时间。借此进行判断。
注意:只要17点前被开始服务,即使17点之后结束的也应该正常输出时间的
#include <cstdio>#include <algorithm>#include <queue>#include <vector>using namespace std;int n, m, k, q;struct CustomerQueue{int startTime;int popTime;queue<int> serviceTime;};int main(){int index=1;scanf("%d%d%d%d", &n, &m, &k, &q);vector<CustomerQueue> windows(n+1);int* serviceTime = new int[k+1];int* finishTime = new int[k+1];vector<bool> sorry(k+1,false);//添加一个存储超时信息的数组,光靠时间判断会出现问题int* query = new int[q+1];for(int i=1; i<=n; i++){//初始化一下,否则在VS下int会初始化为0xccccccccwindows[i].popTime = 0;windows[i].startTime = 0;}for(int i=1; i<=k; i++){scanf("%d", &serviceTime[i]);}for(int i=1; i<=m; i++){for(int j=1;j<=n; j++){if(index<=k){windows[j].serviceTime.push(serviceTime[index]);if(windows[j].startTime>=540)sorry[index]=true;windows[j].startTime += serviceTime[index];finishTime[index] = windows[j].startTime;if(i==1)windows[j].popTime = windows[j].startTime;index++; }}}while(index<=k){int tmpTime=windows[1].popTime, tmpIndex=1;for(int i=2; i<=n; i++){if(windows[i].popTime<tmpTime){tmpTime = windows[i].popTime;tmpIndex = i;}}windows[tmpIndex].serviceTime.pop();windows[tmpIndex].serviceTime.push(serviceTime[index]);windows[tmpIndex].popTime += windows[tmpIndex].serviceTime.front();if(windows[tmpIndex].startTime>=540)sorry[index]=true;windows[tmpIndex].startTime += serviceTime[index];finishTime[index] = windows[tmpIndex].startTime;index++;}for(int i=1; i<=q; i++){scanf("%d", &query[i]);}for(int i=1; i<=q; i++){if(sorry[query[i]]==false){printf("%02d:%02d\n", (finishTime[query[i]]+480)/60, (finishTime[query[i]]+480)%60);}else{printf("Sorry\n");}}return 0;}
PS. 特别注意一下初始化的问题,如int全局变量会初始化为0,但是在struct中的局部变量就不会,要注意!
- PAT甲级练习1014. Waiting in Line (30)
- 【PAT甲级】1014. Waiting in Line (30)
- PAT甲级1014. Waiting in Line (30)
- 1014. Waiting in Line (30) PAT 甲级
- PAT 甲级 1014. Waiting in Line (30)
- PAT 甲级 1014. Waiting in Line
- [PAT-甲级]1014.Waiting in Line
- PAT甲级A1014. Waiting in Line (30)
- 1014. Waiting in Line (30)-PAT甲级真题(queue的应用)
- [PAT甲级]1014. Waiting in Line (30)(银行排队办理业务结束时间 队列的应用)
- [ZJU.PAT] 1014. Waiting in Line (30)
- 1014. Waiting in Line (30)-PAT
- pat 1014. Waiting in Line (30)
- PAT 1014. Waiting in Line (30)
- PAT A 1014. Waiting in Line (30)
- PAT (Advanced) 1014. Waiting in Line (30)
- PAT 1014. Waiting in Line (30)
- PAT 1014. Waiting in Line (30)
- JDK 源码解析 —— Integer 来自 http://blog.csdn.net/wenniuwuren
- 树莓派的一些论坛
- Java实现死锁
- 再学01背包(一)
- 卡通画(矢量风格画)特效生成算法
- PAT甲级练习1014. Waiting in Line (30)
- c++命名空间
- 微信小程序 开发过程中遇到的坑(一)
- systemverilog语法(九)
- 图解Linux命令之--insmod命令
- Lucene初探之总体架构
- 数据可视化
- AndroidStudio插件 - GsonFormat快速实现Javabean
- 绝对路径相对路径