[PAT]1014 Waiting in Line

[C++]1014 Waiting in Line

1014 Waiting in Line:
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 customers 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. Customer₃will 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.

输入格式：
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.

输出格式：
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.

输入：
2 2 7 5
1 2 6 4 3 534 2
3 4 5 6 7
输出：
08:07
08:06
08:10
17:00
Sorry

题目大意：
有n个窗口可排队，每队最多排m个人，k_i表示第i个人的窗口处理时间。
从1~k开始排队，哪个队伍人少就排哪对，如果有队伍人数一样多，则排窗口号数小的队伍。
一个窗口必须要处理完一个人才能继续处理下一个人
问q个人处理完成的时间，如果结束时间超过下午五点，则不能处理，输出sorry。

解题分析：
只有mn个人能排队，mn之后的人需要在黄线外等待，有人处理完之后才能进队
定义结构体，每个队伍保存队头处理的时间poptime和队尾结束处理的时间endtime
找出poptime最小的队伍处理，处理完之后可从黄线后进一人到该队伍队尾，并更新poptime和endtime，重复上述
如果队尾结束时间超过下午五点，则该队伍之后进来的所有人都不能处理完

AC代码：

#include<iostream>
#include<queue>
#include<vector>
using namespace std;

int n, m, k ,q;

struct ST{
	int poptime;
	int endtime;
	queue<int> que;
};

ST wins[25];
int times[1005];
int qs[1005];
int res[1005];
int sorry[1005];

int main(){
	
	cin>>n>>m>>k>>q;
	for(int i = 1; i<=k; i++){
		cin>>times[i];
	}
	
	int index = 1;
	for(int i = 1; i<=m; i++){
		for(int j = 1; j<=n; j++){
			if(index <= k){
				wins[j].que.push(times[index]);
				if(wins[j].endtime >= 540)
					sorry[index] = 1;
				wins[j].endtime += times[index];
				res[index] = wins[j].endtime;
				if(i == 1)
					wins[j].poptime = wins[j].endtime;
				index++; 
			}
		}
	}
	
	while(index <= k) {
		int mintime = wins[1].poptime;
		int mink = 1;
		for(int i = 2; i<=n; i++){
			if(wins[i].poptime < mintime){
				mintime = wins[i].poptime;
				mink = i;
			}
		}
		
		wins[mink].que.pop();
		wins[mink].que.push(times[index]);
		wins[mink].poptime += wins[mink].que.front();
		if(wins[mink].endtime >= 540)
			sorry[index] = 1;
		wins[mink].endtime += times[index];
		res[index] = wins[mink].endtime;
		index++;
	}
	
	for(int i = 1; i<=q; i++){
		int qs;
		cin>>qs;
		if(sorry[qs] == 1)
			cout<<"Sorry"<<endl;
		else 
			printf("%02d:%02d\n", (res[qs]+480)/60, (res[qs]+480)%60); 
	}
	
	return 0;
}