1014 Waiting in Line （30 分)

2019-03-09

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 (N**M+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 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, custome**r1 is served at windo**w1 while custome**r2 is served at windo**w2. Custome**r3 will wait in front of windo**w1 and custome**r4 will wait in front of windo**w2. Custome**r5 will wait behind the yellow line.

At 08:01, custome**r1 is done and custome**r5 enters the line in front of windo**w1 since that line seems shorter now. Custome**r2 will leave at 08:02, custome**r4 at 08:06, custome**r3 at 08:07, and finally custome**r5 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 Sorryinstead.

Sample Input:

1
2
3

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

作者: CHEN, Yue

单位: 浙江大学

时间限制: 400 ms

内存限制: 64 MB

代码长度限制: 16 KB

题目大意

一家银行有n个窗口，每个窗口有一个黄线，黄线之内可以排m位顾客。顾客总是会选择最短的（不是最早结束的队伍）且编号最小的队伍。给出窗口数，黄线内可排队数，顾客数，顾客到达的时间以及处理时间，要求输出顾客结束服务的时间。注意银行在下午5点关门，这个时间之后顾客没有开始被服务，则需输出sorry。

分析

利用结构体模拟队列，记录该队列结束服务的时间以及队伍长度及排队顾客。每次选择最短的队伍，让顾客入队。若队长为m，说明所有队伍已满，选择最早结束的队伍出队，出队时队首顾客结束时间即为队列原结束时间加上其处理时间。最后将所有队伍中的顾客都出队，记录顾客结束时间。

代码

#include <iostream>
#include <vector>
#include <algorithm>
#include <deque>
#include <map>
#include <cstring>
#include <cmath>

using namespace std;

struct  line{
    int endtime=0;
    deque<int> q={};
};
int main() {
    //std::ios::sync_with_stdio(false);
    int choice;

    int n, m, k, q, out, cur;
    int pro[1010] = {}, fin[1010] = {};
    cin >> n >> m >> k >> q;
    for (int i = 1; i <= k; i++)
        cin >> pro[i];
    line lines[n];
    for (int i = 1; i <= k; i++) {  
        int minl = m + 1;
        int mint = 9999999;
        for (int j = 0; j < n; j++) {
            line l = lines[j];
            int ll = l.q.size();
            if (minl > ll) {
                choice = j;
                minl = l.q.size();
            }
        }
        if (minl == m) {
            for (int j = 0; j < n; j++) {
                line l = lines[j];
                if (mint > l.endtime + pro[l.q.front()]) {
                    mint = l.endtime + pro[l.q.front()];
                    choice = j;
                }
            }
        }

        if (lines[choice].q.size() == m) {
            out = lines[choice].q.front();
            lines[choice].q.pop_front();
            fin[out] = lines[choice].endtime + pro[out];
            lines[choice].endtime += pro[out];
        }
        lines[choice].q.push_back(i);


    }
    for (int i = 0; i < n; i++) {
        while (!lines[i].q.empty()) {
            int out = lines[i].q.front();
            fin[out] = lines[i].endtime + pro[out];
            lines[i].endtime += pro[out];
            lines[i].q.pop_front();
        }
    }
    for (int i = 0; i < q; i++) {
        cin >> cur;
        int hh = 8 + fin[cur] / 60;
        int mm = fin[cur] % 60;
        if (fin[cur] <= 540 || (fin[cur] - pro[cur]) < 540)
            printf("%02d:%02d\n", hh, mm);
        else
            printf("Sorry\n");

    }
}