PAT 1014

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 

---

最大的坑是：对于那些在17:00之前已经开始处理的客户必须将他们的事务处理完。

代码

  1 #include <stdio.h>
  2 #include <string.h>
  3 
  4 typedef struct Queue{
  5     int q[11];
  6     int s,e;
  7 }Queue;
  8 void print(int);
  9 void initQueue(Queue *);
 10 int isEmpty(Queue *);
 11 int isFull(Queue *);
 12 int enterQueue(Queue *,int);
 13 int outQueue(Queue *);
 14 int readQueueBase(Queue *);
 15 
 16 int processTime[1004],remindedTime[1004],queries[1000];
 17 int finishedTime[1004];
 18 Queue queue[20];
 19 const int totalTime = 540;
 20 int main()
 21 {
 22     int N,M,K,Q;
 23     int i,j;
 24     while(scanf("%d%d%d%d",&N,&M,&K,&Q) != EOF){
 25         for(i=1;i<=K;++i){
 26             scanf("%d",&processTime[i]);
 27             remindedTime[i] = processTime[i];
 28         }
 29         for(i=0;i<Q;++i)
 30             scanf("%d",&queries[i]);
 31         memset(finishedTime,0,sizeof(finishedTime));
 32         for(i=0;i<N;++i)
 33             initQueue(&queue[i]);
 34         int yellowLineNum = 1;
 35         for(i=0;i<M;++i){
 36             for(j=0;j<N;++j){
 37                 if(yellowLineNum <= K){
 38                     enterQueue(&queue[j],yellowLineNum);
 39                     ++yellowLineNum;
 40                 }
 41                 else
 42                     break;
 43             }
 44             if(yellowLineNum > K)
 45                 break;
 46         }
 47         int nowTime = 1;
 48         int x;
 49         for(;nowTime <= totalTime;++nowTime){
 50             for(i=0;i<N;++i){
 51                 if(!isEmpty(&queue[i])){
 52                     x = readQueueBase(&queue[i]);
 53                     --remindedTime[x];
 54                     if(remindedTime[x] == 0){
 55                         finishedTime[x] = nowTime;
 56                         outQueue(&queue[i]);
 57                         if(yellowLineNum <= K){
 58                             enterQueue(&queue[i],yellowLineNum);
 59                             ++yellowLineNum;
 60                         }
 61                     }
 62                 }
 63             }
 64         }
 65         for(i=0;i<N;++i){
 66             if(!isEmpty(&queue[i])){
 67                 x = readQueueBase(&queue[i]);
 68                 if(remindedTime[x] < processTime[x])
 69                     finishedTime[x] = totalTime + remindedTime[x];
 70             }
 71         }
 72         for(i=0;i<Q;++i){
 73             if(finishedTime[queries[i]])
 74                 print(finishedTime[queries[i]]);
 75             else
 76                 printf("Sorry\n");
 77         }
 78     }
 79     return 0;
 80 }
 81  
 82 void print(int t)
 83 {
 84     int h = t / 60;
 85     int s = t % 60;
 86     printf("%02d:%02d\n",h+8,s);
 87 }
 88 
 89 void initQueue(Queue *Q)
 90 {
 91     (*Q).s = (*Q).e = 0;
 92 }
 93 
 94 int isEmpty(Queue *Q)
 95 {
 96     return ((*Q).s) == ((*Q).e); 
 97 }
 98 
 99 int isFull(Queue *Q)
100 {
101     return ((*Q).e + 1) % 11 == ((*Q).s);
102 }
103 
104 int enterQueue(Queue *Q,int x)
105 {
106     (*Q).q[(*Q).e] = x;
107     (*Q).e = ((*Q).e + 1) % 11;
108     return 1;
109 }
110 
111 int outQueue(Queue *Q)
112 {
113     int x = (*Q).q[(*Q).s];
114     (*Q).s = ((*Q).s + 1) % 11;
115     return x;
116 }
117 
118 int readQueueBase(Queue *Q)
119 {
120     return (*Q).q[(*Q).s];
121 }