题目 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:
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 custome2 is served at window2. Customer 3 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
n
个窗口,每个窗口前可以排队m
人。k
位用户需要服务,给出了每位用户服务需要占用的的minute
数。8
点开始服务,如果有窗口还没排满就在窗口前排队,否则就在黄线外等候。如果有某一队列有一个用户走了服务完毕了,黄线外的人就进来一个。如果同时有多个窗口走了一个人,就选窗口数最小的那个窗口前去排队。q
个人,以HH:mm
的格式输出他们的服务结束时间。银行在17:00
点停止服务,如果一个客户在17:00
以及以后还没有开始服务(此处不是结束服务是开始17:00
)就输出Sorry
。这个题最易错的地方是如何判断一个客户无法完成服务,
是看他的结束时间是否超过17:00
吗??不不不不不,只要他在17:00前开始被服务,银行就必须给他服务完才结束下班。所以我们要看他的开始服务时间是不是超过17:00
,而不是结束时间!!!
队列
了,每个窗口前都有一个队,假如有n个窗口,那就建n个队列呗,但是因为编号都是从1开始,所以直接建n+1个队列好了8*60
,17:00就相当于17*60
,最后输出HH:mm很好办printf("%02d:%02d\n", customer[id].end_time / 60, customer[id].end_time % 60);
n*m
个人可以直接排到黄线前面,所以这部分人要单独处理,后面的人才需要等前面有人走了再去排队。个人觉得我代码注释够清楚了,应该都能看懂的吧,哈哈。
#include <iostream>
#include <queue>
using namespace std;
struct Customer {
int cost_time, start_time, end_time;
} customer[1001];
int main() {
// n个窗口,每个窗口到黄线,最多排m个人,k个客户,有q个客户想知道他啥时候被服务结束
int n, m, k, q;
cin >> n >> m >> k >> q;
// 银行开始服务时间,结束服务时间
int open_time = 8 * 60, close_time = 17 * 60;
// n个窗口,从1开始
queue<Customer> windows[n + 1];
// 读入k个人花费的时间
for (int i = 1; i <= k; ++i)
cin >> customer[i].cost_time;
// 排队
for (int i = 1; i <= k; ++i) {
// 黄线内能排 n * m个人
if (i <= n * m) {
// 按顺序挨个窗口,顾客id和窗口id都从1开始
int window = (i - 1) % n + 1;
// 前n个人,一人一个窗口
if (i <= n) {
// 开始时间是银行开始服务时间
customer[i].start_time = open_time;
// 他的结束时间是他的开始加+他的服务时间
customer[i].end_time = open_time + customer[i].cost_time;
} else {
Customer last_customer = windows[window].back();
// 他的开始时间是上一个人的结束时间
customer[i].start_time = last_customer.end_time;
// 他的结束时间是他的开始加+他的服务时间
customer[i].end_time = customer[i].start_time + customer[i].cost_time;
}
// 排到对应窗口
windows[window].push(customer[i]);
// 黄线外的人
} else {
// 看哪一个窗口先离开一个人,就是看每个窗口前那个队伍中第一个人的离开时间
int window = 1;
for (int j = 2; j <= n; ++j) {
// 除非第一个人离开的更早,才选择这个窗口
if (windows[j].front().end_time <
windows[window].front().end_time) {
window = j;
}
}
// 这个窗口第一个人出队
windows[window].pop();
// 排到这个窗口
Customer last_customer = windows[window].back();
// 他的开始时间是上一个人的结束时间
customer[i].start_time = last_customer.end_time;
// 他的结束时间是他的开始加+他的服务时间
customer[i].end_time = customer[i].start_time + customer[i].cost_time;
//入队
windows[window].push(customer[i]);
}
}
// q个人想要查询自己什么时间能结束
int id;
while (q-- > 0) {
cin >> id;
// 开始服务时间超过了银行关门时间,无法服务
if (customer[id].start_time >= close_time)
cout << "Sorry" << endl;
else
printf("%02d:%02d\n", customer[id].end_time / 60,
customer[id].end_time % 60);
}
return 0;
}