题目
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.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, customer?1?? is served at window?? while customer?2?? is served at window?2. Customer3 will wait in front of window??? and customer?4 will wait in front of window?2?? . Customer?5?? will wait behind the yellow line.
At 08:01, customer1??? is done and customer?5?? enters the line in front of window?1?? since that line seems shorter now. Customer?2?? will leave at 08:02, customer?4?? at 08:06, customer?3 at 08:07, and finally customer?5 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
思路
这是一道模拟题，模拟一下排队问题。
只在黄线内排队，黄线外的要等有人完成时采取排队，服务开始时间必须早与17点，否则是不能办理。
最简单的想法是，先填满m?nm*nm?n这个队列，然后走一个进一个。需要考虑人不多的情况，即人数不够m?nm*nm?n时的情况。时间直接用分钟为单位，最后再转换为指定的格式。
#include <stdio.h>
#include <queue>using namespace std;int longest = (17-8)*60;void show(int time,int i){int h ,m ;if(time -i <longest){h = time /60 +8;m = time % 60;printf("%02d:%02d\n",h,m);}else{printf("Sorry\n");}
}int main(){int n,m,k,q,i,j,flag = 0,count = 0 ,next,t;if(4 != scanf("%d %d %d %d",&n,&m,&k,&q))return -2;queue<int> line[n];int timer[k]={0};int over[k]={0};int query[q]={0};for(i=0 ;i<k ;i++){if( 1 != scanf("%d",&(timer[i])))return -2;}for(i=0 ;i<q ;i++){if( 1 != scanf("%d",&(query[i])))return -2;}for(i=0;i<n*m;i++){if(i>=k){flag = 1;break;}next = i%n;t = 0;if(line[next].size() == 0){t = timer[i];}else{t = line[next].back()+timer[i];}line[next].push(t);over[i]= t;}if(!flag){for(;i<k;i++){next = 0;for(j=0;j<n;j++){if(line[next].front() > line[j].front()){next = j;}}line[next].push(line[next].back()+timer[i]);over[i]= line[next].back();line[next].pop();}}for(i=0;i<q;i++){next = query[i]-1;show(over[next],timer[next]);}return 0;
}