1014. Waiting in Line (30)

Suppose a bank has N windows open for service. There is a yellow line in front of the windows which divides 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.

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

算法分析：采用队列的数据结构，每个窗口看成是一个队列，再创建一个结构体存储每个客户需要花费的时间和离开的时间（距离银行开门到这个客户离开的分钟数），还要一个数组来存储每个窗口在服务一个客户之后的时间，具体算法见代码注释。

#include <stdio.h>#include <stdlib.h>#define INF 65535struct custom //每个客户花费的时间和离开的时间{    int serve;    int leave;}cus[1001];typedef struct QNode* queue; //队列struct QNode{    int vertex[1000];    int post, next;};int isEmpty(queue q) //貌似不用判断队列是否为空，习惯性的写上了。{    if(q->next == q->post)        return 1;    else        return 0;}void addQ(queue q, int v){    q->vertex[++q->next] = v;}int deleteQ(queue q){    return q->vertex[++q->post];}int firstCus(queue q) //返回排在最前的客户{    int index = q->post + 1;    return q->vertex[index];}int main(){    queue window[20]; //窗口    int time[20] = {0};     int n, m, k, q, i, j;    scanf("%d %d %d %d", &n, &m, &k, &q);    for(i = 0; i <= n; i++) //初始化窗口队列    {        window[i] = (queue)malloc(sizeof(struct QNode));        window[i]->next = window[i]->post = -1;    }    for(i = 1; i <= k; i++) //初始化客户的时间    {        scanf("%d", &cus[i].serve);        cus[i].leave = INF;    }    for(i = 1; i <= n * m && i <= k; i++) //站满黄线内的空间，注意当人数过少不用在黄线外等    {        cus[i].leave = time[i % n] + cus[i].serve; //更新客户离开的时间        //用窗口服务的时长+客户花费的时间        time[i % n] = cus[i].leave;         addQ(window[i % n], i);    }    for( ; i <= k; i++) //剩下的人从黄线外坐等    {        int mint = INF, index = -1;        for(j = 0; j < n; j++) //选出最先结束的人        {            int first = firstCus(window[j]);            if(cus[first].leave < mint)            {                index = j;                mint = cus[first].leave;            }        }        cus[i].leave = time[index] + cus[i].serve;        time[index] = cus[i].leave;        deleteQ(window[index]); //完事后离开队列        addQ(window[index], i); //在黄线外的人进入队列    }    for(i = 0; i < q; i++) //输出    {        int c;        scanf("%d", &c);        if(cus[c].leave - cus[c].serve < 540)            printf("%02d:%02d\n", 8 + cus[c].leave / 60, cus[c].leave % 60);        else            printf("Sorry\n");    }    return 0;}