PAT 甲级 1014. Waiting in Line

原题传送门

队列queue的应用
判断是否出队的方法：模拟时间流动，依次检查各个窗口队列
时间的处理：先用十进制，最后输出时间格式

#include<iostream>
#include<queue>
using namespace std;const int MAXN = 22;
const int MAXK = 1000 + 5;
int N, M, K, Q; // 窗口数、黄线内容量、顾客数量、查询数量
int now_time = 0;
int close_time = (17 - 8) * 60;struct customer {int id;int process_time;   // 服务所需时间int done_time;  // 服务完成时间customer():done_time(-1) {}  // done_time=-1没有更新时输出sorry
} customers[MAXK];queue<customer> line[MAXN], behindQueue;    // 每个窗口对应一个黄线内队列line，黄线外是一个排队队列
int last_time[MAXN] = { 0 };    // 记录每个窗口队列的前一次的完成时间，即队头的开始时间int main() {cin >> N >> M >> K >> Q;int capacity = N * M;for (int i = 1; i <= K; i++) {cin >> customers[i].process_time;customers[i].id = i;if (i <= capacity) {line[(i - 1) % N + 1].push(customers[i]);} else {behindQueue.push(customers[i]); // 黄线外排队}}int count = K;while (now_time<close_time && count) {   // 模拟时间流动，每分钟检查一次有没有完成的now_time++;for (int i = 1; i <= N; i++) {  // 依次对各窗口的队头进行检查if (!line[i].empty()) {int done_time = line[i].front().process_time + last_time[i];    // 队头客户的完成时间int id = line[i].front().id;    // 队头客户的idif (last_time[i]<now_time && (done_time>close_time || done_time==now_time)) {   // 这样理解：在前一个人完成时间小于现在时间的情况下（即在17点前开始服务），队头超时完成服务或者现在完成了服务，这时候正式更新done_timecustomers[id].done_time = done_time;last_time[i] = done_time;line[i].pop();count--;if (behindQueue.size()) {line[i].push(behindQueue.front());behindQueue.pop();}} else if (last_time[i] >= now_time) {  // 17点前还没开始服务的，黄线内客户直接跳过，不更新done_time）line[i].pop();count--;if (behindQueue.size()) {line[i].push(behindQueue.front());behindQueue.pop();}}}}}int query;for (int i = 0; i < Q; i++) {   // 输出各个有效的done_timecin >> query;if (customers[query].done_time != -1)printf("%02d:%02d\n", 8 + customers[query].done_time / 60, customers[query].done_time % 60);elseprintf("Sorry\n");}return 0;
}

附原题：

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

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