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
Submit01:
#include <iostream> #include <algorithm> #include <queue> using namespace std; #define maxK 1001//最多有1001个人 #define maxN 20//n最大20个窗口 #define maxM 10//m黄线前最多可以排10个人 struct Window{ queue<int> q;//总人数队列 int popTime;//每个人所需服务时间 }windows[maxN]; int cost[maxK]; int finished[maxK]; int n,m,k,q; //每个窗口黄线位置填充 int fillWindow() { int id = 1; for (int j=0;j<n;j++) { windows[j].popTime = cost[id]; windows[j].q.push(id); if(++id > k) return id; } for (int i=0;i<m-1;i++) { for (int j=0;j<n;j++) { windows[j].q.push(id); if(++id > k) return id; } } return id; } int main() { scanf("%d %d %d %d",&n,&m,&k,&q); for (int i=1;i<=k;i++) { scanf("%d",&cost[i]); } fill(finished,finished+maxK,-1);//初始化完成时间，-1表示无法完成任务 int id= fillWindow(); for (int t=0;t<540;t++) { for (int i=0;i<n;i++) { if (windows[i].popTime != t) continue; finished[windows[i].q.front()] = t;//记录完成时间 windows[i].q.pop();//出列 if(windows[i].q.size() > 0) { windows[i].popTime = t + cost[windows[i].q.front()];//更新出列时间 } if (id > k) continue;// windows[i].q.push(id++);//出列一个黄线外客户进入一个 } } for (int i=0; i<n;i++) { if(windows[i].q.size() < 1) continue; finished[windows[i].q.front()] = windows[i].popTime;//窗口有用户则完成时间更新，否则该用户无法完成任务 } int query, timestamp; for (int i=0;i<q;i++) { scanf("%d", &query); timestamp = finished[query];//完成时间赋值，=-1则输出sorry if (timestamp == -1) { printf("Sorry "); continue; } printf("%02d:%02d ",timestamp / 60 + 8, timestamp % 60);//小时，分钟 } return 0; }
Submit02:
#include <iostream> #include <queue> #include <vector> using namespace std; struct node { int poptime, endtime;//poptime-标记最小的那个人最先服务完，endtime-服务完成时间判断是否sorry queue<int> q; }; int main() { int n, m, k, q, index = 1;//n个窗口，每个窗口可以有m个人排队，如果满了，就在黄线外，总共k个人，目标：q个人的服务时间 scanf("%d%d%d%d", &n, &m, &k, &q); vector<int> time(k + 1), result(k + 1); for(int i = 1; i <= k; i++) { scanf("%d", &time[i]);//每个用户需要服务的时间 minute } vector<node> window(n + 1); vector<bool> sorry(k + 1, false); for(int i = 1; i <= m; i++) { for(int j = 1; j <= n; j++) { if(index <= k) { window[j].q.push(time[index]); if(window[j].endtime >= 540) sorry[index] = true; window[j].endtime += time[index]; if(i == 1) window[j].poptime = window[j].endtime; result[index++] = window[j].endtime; } } } while(index <= k) { int tempmin = window[1].poptime, tempwindow = 1; for(int i = 2; i <= n; i++) { if(window[i].poptime < tempmin) { tempwindow = i; tempmin = window[i].poptime; } } window[tempwindow].q.pop(); window[tempwindow].q.push(time[index]); window[tempwindow].poptime += window[tempwindow].q.front(); if(window[tempwindow].endtime >= 540) sorry[index] = true; window[tempwindow].endtime += time[index]; result[index++] = window[tempwindow].endtime; } for(int i = 1; i <= q; i++) { int query, minute; scanf("%d", &query); minute = result[query]; if(sorry[query] == true) printf("Sorry "); else printf("%02d:%02d ",(minute + 480) / 60, (minute + 480) % 60); } return 0; }
参考：
柳婼-https://blog.csdn.net/liuchuo/article/details/54561626
昵称五个字-https://blog.csdn.net/a617976080/article/details/89676670