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

思路：

 本题涉及到时间换算差值的问题，先将所有时间化到 秒，这样方便计算

符合要求的客户存入vector中，因为不知道符合要求的客户有多少位，就不用再去计数

三个窗口服务：对于每一个客户，找一个结束时间最早的窗口服务

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;

const int K = 111;
const int INF = 1000000000;
struct Customer{
    int comeTime, serveTime;
}newCustomer;
vector<Customer> custom;
int endTime[K];     //记录每个窗口的结束时间

int convertTime(int h, int m, int s){
    return h * 60 * 60 + m * 60 + s;
}

bool cmp(Customer a, Customer b){
    return a.comeTime < b.comeTime;
}

int main(){
    int c, w, totTime = 0;
    int stTime = convertTime(8, 0, 0);
    int edTime = convertTime(17, 0, 0);
    scanf("%d %d", &c, &w);
    for(int i = 0; i < w; i++){
        endTime[i] = stTime;
    }
    for(int i = 0; i < c; i++){
        int h, m, s, serveTime;
        scanf("%d:%d:%d %d", &h, &m, &s, &serveTime);
        int comeTime = convertTime(h, m, s);
        if(comeTime > edTime)
            continue;
        newCustomer.comeTime = comeTime;
        newCustomer.serveTime = serveTime <= 60 ? serveTime * 60 : 3600;
        custom.push_back(newCustomer);
    }
    sort(custom.begin(), custom.end(), cmp);
    for(int i = 0; i < custom.size(); i++){
        int idx = -1, minEndTime = INF;
        for(int j = 0; j < w; j++){
            if(endTime[j] < minEndTime){
                idx = j;
                minEndTime = endTime[j];
            }
        }
        if(endTime[idx] <= custom[i].comeTime){
            endTime[idx] = custom[i].comeTime + custom[i].serveTime;
        }
        else{
            totTime += (endTime[idx] - custom[i].comeTime);
            endTime[idx] += custom[i].serveTime;
        }
    }
    if(custom.size() == 0)
        printf("0.0\n");
    else
        printf("%.1f", totTime / 60.0 / custom.size());
    return 0;
}