Suppose a bank has K windows open for service. There is a yellow line in front of the windows which devides the waiting area into two parts. All the customers have to wait in line behind the yellow line, until it is his/her turn to be served and there is a window available. It is assumed that no window can be occupied by a single customer for more than 1 hour.

Now given the arriving time T and the processing time P of each customer, you are supposed to tell the average waiting time of all the customers.

Input Specification:
Each input file contains one test case. For each case, the first line contains 2 numbers: N (≤$10^4$) - the total number of customers, and K (≤100) - the number of windows. Then N lines follow, each contains 2 times: HH:MM:SS - the arriving time, and P - the processing time in minutes of a customer. Here HH is in the range [00, 23], MM and SS are both in [00, 59]. It is assumed that no two customers arrives at the same time.

Notice that the bank opens from 08:00 to 17:00. Anyone arrives early will have to wait in line till 08:00, and anyone comes too late (at or after 17:00:01) will not be served nor counted into the average.

Output Specification:
For each test case, print in one line the average waiting time of all the customers, in minutes and accurate up to 1 decimal place.

Sample Input:

7 3
07:55:00 16
17:00:01 2
07:59:59 15
08:01:00 60
08:00:00 30
08:00:02 2
08:03:00 10

Sample Output:

8.2

解題思路：
利用優先佇列模擬。

#include <iostream>
#include<cstdio>
#include<queue> 
#include<algorithm>
#include<cmath>
#define ON 8*60*60//開始時間
#define OFF 17*60*60//不能超過的時間
using namespace std;
int num(char c){
	return c-'0';
}
struct node{
	int start,last,window,time,wait;//開始時間，最後執行完該顧客需求時的時間，顧客所在視窗，顧客需求時間，等待時間
	int h,m,s;
}cus[10010];
int N,K;
bool cmp1(node &a,node &b){//按出現時間排序
	if(a.h!=b.h) return a.h<b.h;
	if(a.m!=b.m) return a.m<b.m;
	if(a.s!=b.s) return a.s<b.s;
	return a.time<b.time;
}

struct cmp{//利用優先佇列求出當前最快結束的視窗
	bool operator()(node a,node b){
		if(a.last!=b.last)
			return a.last>b.last;//降序 
		return a.window>b.window;
	}
}; 

priority_queue<node,vector<node>,cmp>Q;
int cnt[110];
int main(int argc, char** argv) {
	scanf("%d%d",&N,&K);
	for(int i=0;i<N;i++){
		char input[10];
		scanf("%s%d",input,&cus[i].time);
		cus[i].time*=60;//記得全部化成秒
		cus[i].h=num(input[0])*10+num(input[1]);
		cus[i].m=num(input[3])*10+num(input[4]);
		cus[i].s=num(input[6])*10+num(input[7]);
		cus[i].start=cus[i].h*60*60+cus[i].m*60+cus[i].s;//s 
		if(cus[i].h>17||(cus[i].h==17&&(cus[i].m>0||cus[i].s>0))){//超過17點，捨棄
			i--;
			N--;
		}
	}
	sort(cus,cus+N,cmp1);

	for(int i=0;i<K;i++){//利用cnt[i]表示第i個視窗最後結束所需要的時間
		cnt[i]=ON;
	}
	int sum=0,ind=-1;
	for(int i=0;i<N;i++){
		if(Q.size()<K){
			ind=cus[i].window=i%K;
			cus[i].last=cus[i].time+max(cnt[ind],cus[i].start);//這個地方有可能出現的時間晚於前i個視窗執行的時間，中間有空餘；第二個情況是出現的時間早於8點，均能處理
			cus[i].wait=cus[i].last-cus[i].time-cus[i].start;//第i個顧客的等待時間
			Q.push(cus[i]);
			cnt[ind]=cus[i].last;//更新每個視窗最後的時間
		}
		else{
			node tmp=Q.top();
			Q.pop();
			ind=cus[i].window=tmp.window;	
			cus[i].last=cus[i].time+max(cnt[ind],cus[i].start);
			cus[i].wait=cus[i].last-cus[i].time-cus[i].start;
			Q.push(cus[i]);
			cnt[ind]=cus[i].last; 
		}
	}
	
	int count=0;
	for(int i=0;i<N;i++){
//		if(cus[i].last<=OFF){//超過17點？？ 這個地方表示最後的執行時間超過17點也符合題意，只有到達時間超過17點不算等待，因此捨去這種情況。
			count++;
			sum+=cus[i].wait;
//		}
	}
	printf("%.1lf",sum/(60.0*count));
	return 0;
}