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 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.
  • Customeri​ will take Ti​ 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 customers 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 Specification:

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 Specification:

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

 

题目大意:有一个银行里有N个窗口，每个窗口最多能排M个人。客户总是选择人最少的窗口排队，如果没有窗口空余，则继续等待。现在给出每个人处理业务所需的时间，问其在什么时候处理完业务。银行8:00上班，17:00下班，假设所有人都是在8:00就来到银行，另外，如果不能在下班之前处理完业务，则输出Sorry。

解题思路:这是一道比较典型的模拟题。我的想法是首先设立窗口结构体，结构体内包括一个队列(储存在这个窗口队的人所需的时间)q，该窗口可以接收人处理业务的时间（便于选择）poptime，该窗口结束所有业务的时间(判断是否能提供服务)endtime。每次遍历选择一个poptime最小的窗口。使用一个数组判断用户是否能完成业务，用户选作窗口时，如果该窗口的endtime超过下班时间，则在数组中记录这个用户(输出Sorry)。

代码如下:

 1 #include<iostream>
 2 #include<vector>
 3 #include<algorithm>
 4 #include<queue>
 5 using namespace std;
 6 struct node{    // 窗口结构体
 7     int poptime,endtime;
 8     queue<int>q;
 9 };
10 int main()
11 {
12     int n,m,cnt,query,index=1;
13     scanf("%d %d %d %d",&n,&m,&cnt,&query);
14     vector<int>time(cnt+1),res(cnt+1);
15     for(int i=1;i<=cnt;i++)
16         scanf("%d",&time[i]);
17     vector<node>windows(n+1);
18     vector<bool>sorry(cnt+1,false);
19     for(int i=1;i<=m;i++)
20     {
21         for(int j=1;j<=n;j++)
22         {
23             if(index<=cnt)           // 将顾客指派入黄线中
24             {
25                 windows[j].q.push(time[index]);
26                 if(windows[j].endtime>=540)
27                     sorry[index]=true;
28                 windows[j].endtime+=time[index];
29                 if(i==1)
30                     windows[j].poptime = windows[j].endtime;
31                 res[index] = windows[j].endtime;
32                 index++;
33             }
34         }
35     }
36     while(index<=cnt)
37     {
38         int tempmin = windows[1].poptime,tempwin = 1;  //为剩下的顾客查找窗口
39         for(int i=2;i<=n;i++)
40         {
41             if(windows[i].poptime<tempmin)
42             {
43                 tempwin = i;
44                 tempmin = windows[i].poptime;
45             }
46         }
47         windows[tempwin].q.pop();
48         windows[tempwin].q.push(time[index]);
49         windows[tempwin].poptime+=windows[tempwin].q.front();
50         if(windows[tempwin].endtime>=540)
51             sorry[index] = true;
52         windows[tempwin].endtime+=time[index];
53         res[index]+=windows[tempwin].endtime;
54         index++;
55     }
56     for(int i=1;i<=query;i++)
57     {
58         int q,minute;
59         scanf("%d",&q);
60         if(sorry[q])
61             printf("Sorry\n");
62         else
63         {
64             minute = res[q];
65             printf("%02d:%02d\n",(minute+480)/60,(minute+480)%60);
66         }
67     }
68     return 0;
69 }