PAT 三月 08, 2020

PAT甲级 1014 Waiting in Line (30分)

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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 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, customer[1] is served at window[1] while customer[2] is served at window[2]. Customer[3] will wait in front of window[1] and customer[4] will wait in front of window[2]. Customer[5] will wait behind the yellow line.

At 08:01, customer[1] 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

注意点

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服务)就不再服务输出sorry;如果这个服务已经开始了了,无论时间多长都要等他服务完毕。如果当成结束服务,则测试点2,4,5不能通过。

#include <cstdio>
#include <queue>
using namespace std;

const int maxk = 1001, maxn = 21, maxTime = 540;
int n, m, k, q, t[maxk], ans[maxk] = {0};

int main() {
    scanf("%d %d %d %d", &n, &m, &k, &q);
    for (int i = 1; i <= k; i++) {
        scanf("%d", &t[i]);
    }
    queue<int> que[maxn];
    int time = 0, index = 0, query;
    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= n; j++) {
            que[j].push(++index);
            if (index == k) {
                break;
            }
        }
        if (index == k) {
            break;
        }
    }
    while (time != maxTime - 1) {
        time++;
        for (int i = 1; i <= n; i++) {
            if (que[i].size() > 0) {
                int front = que[i].front();
                if (t[front] > 0) {
                    t[front]--;
                    if (t[front] == 0) {
                        que[i].pop();
                        ans[front] = time;
                        if (index != k) {
                            que[i].push(++index);
                        }
                    }
                }
            }
        }
    }
  // 额外处理到17:00还没有服务完的客户
    for (int i = 1; i <= n; i++) {
        if (que[i].size() > 0) {
            int front = que[i].front();
            ans[front] = time + t[front];
        }
    }
    for (int i = 0; i < q; i++) {
        scanf("%d", &query);
        if (ans[query] != 0) {
            printf("%02d:%02d\n", 8 + ans[query] / 60, ans[query] % 60);
        } else {
            printf("Sorry\n");
        }
    }
    return 0;
}
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