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
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1 2 3 |
2 2 7 5 1 2 6 4 3 534 2 3 4 5 6 7 |
Sample Output
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1 2 3 4 5 |
08:07 08:06 08:10 17:00 Sorry |
一道关于多个队列操作的问题，根据题目要求进行判断与操作即可。
然而最近我的审题无力综合症又犯了/(ㄒoㄒ)/~~，又一次看错题了
题目中提到
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.
没错，银行17:00后关门，但是会被拒之门外（get “Sorry” reply）的人是在17:00前无法被服务的人……最后针对超时的判断是服务开始时间……看题不仔细的话会出错的……（审题无力综合症_(:зゝ∠)_）
代码如下（写的有点乱(⊙﹏⊙)）：
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#include<stdio.h> #include<stdlib.h> typedef struct node_ { int no; int serve; struct node_ *next; } node; typedef struct queue_ { int num; node *head; node *tail; } queue; int n,m,k,qs,*time_cost,*time_serve,count; queue *q; int print_time(int minute,int used) { int hour; if(minute-used>=540) { printf("Sorry\n"); } else { hour=minute/60+8; minute%=60; printf("%02d:%02d\n",hour,minute); } } int dequeue() { int i,min=0; node *temp; for(i=0;i<n;i++) { if(q[i].num==0) continue; if(q[i].head->serve<q[min].head->serve) { min=i; } } if(min==n||q[min].num==0) { return 1; } if(q[min].head->next==q[min].tail) { q[min].tail=q[min].head; } temp=q[min].head->next; q[min].head->next=temp->next; if(q[min].head->next!=NULL) { q[min].head->serve=q[min].head->next->serve; } free(temp); q[min].num--; count--; return 0; } int enqueue(int no) { int i,min=0; node *temp; for(i=0;i<n;i++) { if(q[i].num<m) { min=i; break; } } for(i=min;i<n;i++) { if(q[i].num<q[min].num) { min=i; } } if(q[min].tail==q[min].head) { temp=(node *)malloc(sizeof(node)); q[min].tail=temp; q[min].head->next=temp; temp->next=NULL; temp->no=no; temp->serve=q[min].head->serve+time_cost[no]; q[min].head->serve=temp->serve; } else { temp=(node *)malloc(sizeof(node)); temp->next=NULL; temp->no=no; temp->serve=q[min].tail->serve+time_cost[no]; q[min].tail->next=temp; q[min].tail=temp; } time_serve[no]=temp->serve; count++; q[min].num++; return 0; } int main() { int i,j,query; node *temp; scanf("%d%d%d%d",&n,&m,&k,&qs); q=(queue *)malloc(n*sizeof(queue)); time_cost=(int *)malloc((k+1)*sizeof(int)); time_serve=(int *)malloc((k+1)*sizeof(int)); for(i=0;i<n;i++) { q[i].num=0; q[i].head=(node *)malloc(sizeof(node)); q[i].head->no=0; q[i].head->serve=0; q[i].head->next=NULL; q[i].tail=q[i].head; } for(i=1;i<=k;i++) { scanf("%d",time_cost+i); } count=0; for(i=1;i<=k;i++) { if(count>=n*m) { dequeue(); } enqueue(i); } for(i=0;i<qs;i++) { scanf("%d",&query); print_time(time_serve[query],time_cost[query]); } } |