[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.

輸入格式：
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.

輸出格式：
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.

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

題目大意：
有n個窗口可排隊，每隊最多排m個人，k_i表示第i個人的窗口處理時間。
從1~k開始排隊，哪個隊伍人少就排哪對，如果有隊伍人數一樣多，則排窗口號數小的隊伍。
一個窗口必須要處理完一個人才能繼續處理下一個人
問q個人處理完成的時間，如果結束時間超過下午五點，則不能處理，輸出sorry。

解題分析：
只有mn個人能排隊，mn之後的人需要在黃線外等待，有人處理完之後才能進隊
定義結構體，每個隊伍保存隊頭處理的時間poptime和隊尾結束處理的時間endtime
找出poptime最小的隊伍處理，處理完之後可從黃線後進一人到該隊伍隊尾，並更新poptime和endtime，重複上述
如果隊尾結束時間超過下午五點，則該隊伍之後進來的所有人都不能處理完

AC代碼：

#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;
}