https://pintia.cn/problem-sets/994805342720868352/problems/994805491530579968

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

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

代碼：

#include <bits/stdc++.h>
using namespace std;

int N, M;

struct Node{
    int time;
    int wait;
};

vector<Node> v;

bool cmp(const Node &a, const Node &b) {
    return a.time < b.time;
}

int main() {
    scanf("%d%d", &N, &M);
    for(int i = 0; i < N; i ++) {
        Node n;
        int h, m, s, t;
        scanf("%d:%d:%d %d", &h, &m, &s, &t);
        n.time = h * 3600 + m * 60 + s;
        n.wait = t * 60;

        if(n.time <= 61200) v.push_back(n);
    }

    vector<int> win(M, 28800);
    //memset(win, 28800, sizeof(win));
    sort(v.begin(), v.end(), cmp);

    //for(int i = 0; i < v.size(); i ++)
        //printf("%d %d\n", v[i].time, v[i].wait);

    double total = 0.0;
    for(int i = 0; i < v.size(); i ++) {
        int temp = 0, minn = win[0];
        for(int j = 1; j < M; j ++) {
            if(win[j] < minn) {
                minn = win[j];
                temp = j;
            }
        }
        if(win[temp] <= v[i].time)
            win[temp] = v[i].time + v[i].wait;
        else {
            total += (win[temp] - v[i].time);
            win[temp] += v[i].wait;
        }
    }

        //cout << total;
    if(v.size() == 0) printf("0.0\n");
    else printf("%.1lf\n", total / 60.0 / v.size());
    return 0;
}

　　把所有時間轉換成秒 把來的人按照來的時間排序 在 17:00:00 之後的排除 更新最快結束的窗口

FH

PAT 甲級 1017 Queueing at Bank