1017. Queueing at Bank (25)

原題連結
相似題目 1014. Waiting in Line (30)
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 (<=10000) - 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

題目大意：

• 銀行排隊辦理業務，有N個人，K個視窗
• 給出N個人的到達銀行時間，個人辦理業務所需要的時間，求平均等待時間
• 銀行視窗08：00開始辦理業務，下午17：00之後到達的人不接受辦理，不算有效的人
• 每個視窗前只能有一個人辦理業務，其他的人均在黃線外等待，有視窗最先空閒（前一個人辦理業務結束），等待的顧客就可以按照到達時間的先後順序去辦理

思路：

• 定義結構體記錄顧客到達時間come，辦理業務所需要時間time
• 為了方便計算，時間全部換成秒來計算
• 統計並篩選所有辦理業務的人，剔除17：00以後到達的人，按照到達時間排序
• 如果顧客到達時間come大於視窗空閒時間windowTime，就可以直接辦理業務，無需等待，否則等待總時間res += (windowTime - come)
• 最終 等待總時間res/60/ 辦理業務有效人數 = 人均等待時間
• 注意res按秒計算，先換成分鐘，再求平均等待時間

程式碼：

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
struct node{
    int come;//到達時間
    int time;//辦理業務所需要時間
};
int cmp(node a, node b){
    return a.come < b.come;
}
int main()
{
    int n, k;//n個人 k個視窗
    scanf("%d %d", &n, &k);
    vector<node> custom;
    for(int i=0; i<n; i++){
        int hh,mm,ss,time;
        scanf("%d:%d:%d %d", &hh, &mm, &ss, &time);
        int cometime = hh*3600 + mm*60 + ss;
        if(cometime > 61200)//顧客來的時間晚於17:00 無效 直接跳過，無法辦理
            continue;
        node temp;
        temp.come = cometime;
        temp.time = time*60;
        custom.push_back(temp);
    }
    sort(custom.begin(), custom.end(), cmp);
    vector<int> windowTime(k, 28800);//28800代表早上八點
    double res = 0.0;
    for(int i=0; i<custom.size(); i++){
        int minWindow=0;//最早結束的視窗 最早結束的視窗時間
        for(int j=1; j<k; j++){
            if(windowTime[minWindow] > windowTime[j]){
                minWindow = j;
            }
        }
        if(windowTime[minWindow] <= custom[i].come){//顧客來的時候就有空閒視窗
            windowTime[minWindow] = custom[i].come + custom[i].time;
        }else{//顧客來的時候需要等待
            res += (windowTime[minWindow] - custom[i].come);//顧客等待時間
            windowTime[minWindow] +=  custom[i].time;//更新視窗空閒時間
        }
    }
    if(custom.size() == 0){//有效人數為0，直接輸出，除以0無意義
        printf("0.0");
    }else{
        printf("%.1f", res/60.0/custom.size());
    }
    return 0;
}