标签：1014 waiting in line pat advanced level practice
题目信息：
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:
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 Input2 2 7 5 1 2 6 4 3 534 2 3 4 5 6 7Sample Output
08:07 08:06 08:10 17:00 Sorry
代码如下：
#include <stdio.h>
#include <memory.h>
#include <iostream>
#include <queue>
using namespace std;
struct tagque
{
int que_time, loc;
};
tagque tque;
queue<tagque> que[21];
int process_time[1001];//每个客户的时间
int que_time[21];//每个队列当前时间
int final_time[1001];//每个客户离开时间
int N, M, K, Q;
int que_op() //寻找一人完成业务离开银行
{
int i, min = 1;
while (min < N && que[min].empty())
++min;
for (i = min + 1; i <= N; ++i)
{
if (que[i].size() > 0 && que[i].front().que_time + que_time[i] < que[min].front().que_time + que_time[min])
{
min = i;
}
}
if (que[min].empty())
{
return false;
}
else
{
tque = que[min].front();
que[min].pop();
final_time[tque.loc] = que_time[min] + tque.que_time;
que_time[min] += tque.que_time;
return true;
}
}
void insert(int val, int loc)//插入一个客户，索引loc
{
int i;
int min = 1;
for (i = 2; i <= N; ++i)
{
if (que[i].size() < M && que[i].size() < que[min].size())
min = i;
}
if (1 == min && que[1].size() == M) //未找到合适位置再处理一人
{
que_op();
insert(val, loc);
}
else
{
tque.que_time = val;
tque.loc = loc;
que[min].push(tque);
}
}
int main()
{
//freopen("d:\\1.txt", "r", stdin);
memset(final_time, 0, sizeof(final_time));
scanf("%d%d%d%d", &N, &M, &K, &Q);
int i, val;
for (i = 1; i <= N; ++i)
{
que_time[i] = 8 * 60;
}
for (i = 1; i <= K; ++i)
{
scanf("%d", &val);
process_time[i] = val;
insert(val, i);
}
while (que_op()) //处理完所有人
continue;
for (i = 1; i <= Q; ++i)
{
scanf("%d", &val);
if (final_time[val] - process_time[val] >= 1020) //17*60 = 1020,17：00之后开始的才Sorry！开始！！！
printf("Sorry\n");
else
printf("%02d:%02d\n", final_time[val] / 60, final_time[val] % 60);
}
return 0;
}1014. Waiting in Line (30)——PAT (Advanced Level) Practise
标签：1014 waiting in line pat advanced level practice
原文地址：http://blog.csdn.net/xianyun2009/article/details/39557535
