1014. Waiting in Line (30)

题目连接：https://www.patest.cn/contests/pat-a-practise/1014

原题如下：

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

这道题目我就是用队列去模拟，感觉需要理顺逻辑关系。自己的代码中冗杂的地方就是只有通过出队才能统计时间。也不想改了……
大致思路如下：
1、将N*M个顾客入队；
2、将剩余的顾客入队（伴随着出队），此时要进行判断哪个窗口的时间最短；
3、将不空的队列进行出队，至此，每位顾客的时间都已经做了统计。
容易出错的地方是要看清楚“如果开始服务的时间大于或等于17：00”，则输出Sorry，而不是结束的时间
代码如下：

#include<stdio.h>
#include<stdlib.h>
#include<stdbool.h>
#define MAXK 1005
int T[MAXK]={0};
int N,M,K,q;
int customer[MAXK];

typedef struct WNode{
    int time;
}Win;

Win Window[MAXK];

typedef struct QNode{
    int front;
    int rear;
    int Maxsize;
    int *Customer;
    int CustomerIn;
}Quenue;

Quenue Q[MAXK];

void CreatQ(int i)
{
    Q[i].front=Q[i].rear=0;
    Q[i].Maxsize=M;
    Q[i].Customer=(int *)malloc(M*(sizeof(int)));
    Q[i].CustomerIn=0;

}

bool IsEmptyQ(int i)
{
    return (Q[i].rear==Q[i].front);
}

int Pop(int i)
{
    if (!IsEmptyQ(i))
    {
        int tmp,out;
        Q[i].front=(Q[i].front+1)%(Q[i].Maxsize);
        tmp=Q[i].Customer[Q[i].front];
        out=Q[i].CustomerIn;
        Q[i].CustomerIn=tmp;

        return out;
    }
}

bool IsFull(int i)
{
    return ((Q[i].rear+1)%(Q[i].Maxsize)==Q[i].front);
}

void Add(int i,int x)
{
    if (!IsFull(i))
    {
        Q[i].rear=(Q[i].rear+1)%(Q[i].Maxsize);
        Q[i].Customer[Q[i].rear]=x;
    }
}

int MinWindowsTime()
{
    int i;
    int Mintime=65535,Mini=0;
    for (i=1;i<=N;i++)
    {
        if (Window[i].time<Mintime)
        {
            Mintime=Window[i].time;
            Mini=i;
        }
    }
    return Mini;
}


int main()
{

    scanf("%d %d %d %d",&N,&M,&K,&q);
    int i,j;
    for (i=1;i<=K;i++)
    {scanf("%d",&T[i]);
     CreatQ(i);
    }

    //第一批顾客,N个窗口
    for (i=1;i<=N*M;i++)
    {
     Window[i].time=480;
        if (i>K)break;
        if (i<=N)
        {Q[i].CustomerIn=i;
        Window[i].time+=T[i];
        }
        else
        {
            j=i%N;
            if (j==0)j=N;
            Add(j,i);
        }
    }
    //剩余的顾客，一共K个顾客
    int c;
    for (;i<=K;i++)  //将每个顾客进行入队处理，先出队（同时统计其时间）再入队
    {
            int t=MinWindowsTime();  //最先有空位的窗口
            if (t!=0)
            {
                if (!IsEmptyQ(t))
                {c=Pop(t);
                customer[c]=Window[t].time;
                Window[t].time+=T[Q[t].CustomerIn];
                Add(t,i);
                }
            }
    }

    for (i=1;i<=N;i++)   // 对于每个窗口，将还在队伍中的顾客进行出队，并统计时间
    {
        while (!IsEmptyQ(i))
        {c=Pop(i);
        customer[c]=Window[i].time;
        Window[i].time+=T[Q[i].CustomerIn];
        }
         if (IsEmptyQ(i)&& Q[i].CustomerIn!=0)
        {
            c=Q[i].CustomerIn;
            customer[c]=Window[i].time;
            Q[i].CustomerIn=0;
            }
        }

    int v,h,m;
    while (q--)
    {
        scanf("%d",&v);
        if (customer[v]-T[v]<1020)
        {h=customer[v]/60;
        m=customer[v]%60;
        printf("%02d:%02d
",h,m);}
        else printf("Sorry
");
    }
    return 0;
}

陈越老师好像讲过这个问题，有时间再把她的思路方法总结下吧。