九度OJ 1326：Waiting in Line（排队） （模拟）

时间限制：1 秒

内存限制：32 兆

特殊判题：否

提交：220

解决：64

题目描述：

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.

输入：

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.

输出：

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.

样例输入：

2 2 7 5
1 2 6 4 3 534 2
3 4 5 6 7

样例输出：

08:07
08:06
08:10
17:00
Sorry

思路：

模拟题，排队题当然数据结构用队列比较好。

用C语言太苦逼了，还要自己写一个队列，还是C++比较好，以后要用C++了。

需要注意的是：即使是17点之前开始办业务，17点之后仍然没有办完就是sorry。

代码：

#include <stdio.h>
#include <stdlib.h> 
 
#define N 21
#define M 11
#define K 1001
 
typedef struct node1 {
    int proc;
    int begin;
} Cust;
     
typedef struct node2 {
    Cust *cust[M];
    int front;
    int rear;
} Queue;
 
int isEmpty(Queue *q)
{
    return (q->front == q->rear);
}   
     
int isFull(Queue *q)
{   
    return (q->front == (q->rear+1)%M);
}
     
int countQueue(Queue *q)
{       
    return (q->rear + M - q->front) % M;
}           
             
void push(Queue *q, Cust *val)
{       
    q->cust[q->rear] = val;
    q->rear = (q->rear+1)%M;
}           
             
Cust *pop(Queue *q)
{       
    int front = q->front;
    q->front = (q->front+1)%M;
    return q->cust[front];
}
 
Cust *top(Queue *q)
{
    return q->cust[q->front];
}
 
int main(void)
{
    int n, m, k, q, i, j;
    Queue *queue[N];
    Cust *cust[K];
    int query[K];
    int maxTime = (17-8)*60;
 
    while (scanf("%d%d%d%d", &n, &m, &k, &q) != EOF)
    {
        for (i=0; i<n; i++)
        {
            queue[i] = (Queue *)malloc(sizeof(Queue));
            queue[i]->front = queue[i]->rear = 0;
        }
        for (i=0; i<k; i++)
        {
            cust[i] = (Cust *)malloc(sizeof(Cust));
            scanf("%d", &cust[i]->proc);
            cust[i]->begin = maxTime+1;
        }
        for (i=0; i<q; i++)
            scanf("%d", &query[i]);
 
        // init state
        int inQueue = 0;
        int outQueue = k;
        for (i=0; i<m; i++)
        {
            for (j=0; j<n; j++)
            {
                inQueue = i*n + j;
                if (inQueue >= k)
                    break;
                if (i == 0)
                    cust[inQueue]->begin = 0;
                push(queue[j], cust[inQueue]);
            }
            if (inQueue >= k)
                break;
        }
        inQueue = (m*n <= k) ? (inQueue+1) : inQueue;
        outQueue = k - inQueue;
        //printf("inQueue=%d, outQueue=%d, k=%d\n", inQueue, outQueue, k);
     
        // simulate the whole process
        int time;
        for (time=0; time<=maxTime; time++)
        {
            for (i=0; i<n; i++)
            {
                if (isEmpty(queue[i]))
                    break;
                Cust *c = top(queue[i]);
                if (c->begin + c->proc == time)
                {
                    pop(queue[i]);
                    if (outQueue > 0)
                    { 
                        push(queue[i], cust[k-outQueue]);
                        outQueue --;
                    }
                    if (!isEmpty(queue[i]))
                    {
                        top(queue[i])->begin = time;
                        //printf("i=%d, time=%d\n", i, time);
                    } 
                }
            }
            if (isEmpty(queue[0]))
                break;
        }   
                 
        for (i=0; i<q; i++) 
        {
            int id = query[i] - 1;
            int endTime = cust[id]->begin + cust[id]->proc;
            if (endTime > maxTime)
                printf("Sorry\n");
            else
                printf("%02d:%02d\n", 8 + endTime/60, endTime%60);
        }
    }
 
    return 0;
}
/**************************************************************
    Problem: 1326
    User: liangrx06
    Language: C
    Result: Accepted
    Time:10 ms
    Memory:916 kb
****************************************************************/