pat 1014 Waiting in Line (30)

1014. Waiting in Line (30)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 51 2 6 4 3 534 23 4 5 6 7

Sample Output

08:0708:0608:1017:00Sorry

这个题是模拟进程的，然而我并不知道，我就是模拟的，用的队列。

最坑爹的地方在最后一句话“ who cannot be served before 17:00。。。”,我一直以为是最后的时间到了就关门，做不完拉倒，然而并不是，当时一直错，心里很纳闷，就从网上找找，结果是自己理解错了，然后又改，又错了，原因是“之前”,有个before没看见，结果又把等号去掉，结果过了。就是用队列模拟，没啥好说的。读题不够认真是真的。

#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>#include<map>#include<queue>using namespace std;struct node{    int x,ord;}a[1010];int main(){    int n,m,k,q;    queue<node>q_line[1010];    int len = 0;    int peo = 0;    int h = 8,ms = 0;    int h_array[1010],m_array[1010];    int s_array[1010],sm_array[1010];    scanf("%d%d%d%d",&n,&m,&k,&q);    for(int i = 1;i<=k||peo<k;i++){        if(i<=k){            scanf("%d",&a[i].x);            a[i].ord = i;        }         if(len == n*m||i>k){            int tmp;            int a_num = 0x3f3f3f,now_number;            for(int j = 0;j<n;j++){                if(q_line[j].empty())continue;                if(q_line[j].front().x < a_num){                    now_number = j;                    a_num  = q_line[j].front().x;                }            }            ms+=a_num;            h+=ms/60;            ms %= 60;            for(int j = 0;j<n;j++){                if(q_line[j].empty())continue;                int ord_temp = q_line[j].front().ord;                if(j == now_number){                   h_array[ord_temp] = h;                   m_array[ord_temp] = ms;                   q_line[j].pop();                   len--;                   peo++;                   if(q_line[j].empty())continue;                   s_array[q_line[j].front().ord] = h;                   sm_array[q_line[j].front().ord] = ms;                }else{                    tmp = q_line[j].front().x - a_num;                    if(tmp)q_line[j].front().x = tmp;                    else {                            q_line[j].pop();                            h_array[ord_temp] = h;                            m_array[ord_temp] = ms;                            len--;                            peo++;                            if(q_line[j].empty())continue;                            s_array[q_line[j].front().ord] = h;                            sm_array[q_line[j].front().ord] = ms;                    }                }            }        }        if(len < n*m && i<=k){            int  a_line = 10010,now_line;            for(int j = 0;j<n;j++){                if(q_line[j].size()<a_line){                    now_line = j;//calcute the shortest line                    a_line = q_line[j].size();                }            }            q_line[now_line].push(a[i]);            s_array[i] = 8;            sm_array[i] = 0;            len++;        }    }    int x;    for(int i = 0 ;i<q;i++){        scanf("%d",&x);        if(s_array[x]*100 + sm_array[x] >= 1700)puts("Sorry");        else{            if(h_array[x]<=9)printf("0%d:",h_array[x]);            else {                    printf("%d:",h_array[x]);            }            if(m_array[x]<=9)printf("0%d\n",m_array[x]);            else {                    printf("%d\n",m_array[x]);            }        }    }}