1017. Queueing at Bank (25)

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

思路：考察的是排队事件的模拟题 涉及到字符串time2整形以及反向变化。对到达的数据小->大排序（必定是最早到的最先得到服务），对窗口进行筛选，因为每次必定是先找最小时间的窗口进行服务的！

#include <iostream>
#include <string>
#include <iomanip>
#include <algorithm>
#include <stdio.h>
using namespace std;
struct Node{
    string arrive;
    int process;
};
int time2int(string s){
    int base = 3600;
    int time = 0;
    int loc;
    for(int i=0;i<3;i++){
        loc = s.find(":");
        time += atoi(s.substr(0,loc).c_str())*base; //This part is the core
        s = s.substr(loc+1);
        base /= 60;
    }
    return time;
}
string int2time(int n){
    string s;
    char part[3];
    for(int i=0;i<3;i++){
        sprintf(part,"%02d",n%60);  //change the int to char[]
        s = (string)part +s;
        n /= 60;
        if(i!=2)
            s = ":"+s;
    }
    return s;
}
bool compare(Node a,Node b){
    return a.arrive<b.arrive;
}
int main()
{
    int N,K;
    cin>>N>>K;
    Node *arr = new Node[N];
    int *waitTime = new int[N];
    string *window = new string[K];
    for(int i=0;i<K;i++){
        window[i] = "08:00:00";
    }
    for(int i=0;i<N;i++){
        cin>>arr[i].arrive>>arr[i].process;
    }
    sort(arr,arr+N,compare);    //Finish sort the arr
    int minN = 0;
    int flag = 0;   //mark the available time
    for(int i=0;i<N;i++){
        if(time2int(arr[i].arrive)>17*3600){
            continue;
        }
        for(int j=0;j<K;j++){   //find the min window
            if(window[j]<window[minN]){
                minN = j;
            }
        }
        if(arr[i].arrive<window[minN]){ //you need to wait
            int windowTime;
            waitTime[i] = time2int(window[minN])-time2int(arr[i].arrive);  //it is by seconds
//            cout<<waitTime[i]<<endl;
            windowTime = time2int(window[minN]) + arr[i].process*60;
            window[minN] = int2time(windowTime);
            flag++;
        }else{
            int windowTime;
            waitTime[i] = 0;
//            cout<<waitTime[i]<<endl;
            windowTime = time2int(arr[i].arrive) + arr[i].process*60;
            window[minN] = int2time(windowTime);
            flag++;
        }
    }
    double result = 0.0;
    for(int i=0;i<flag;i++){
        result += waitTime[i];
    }
    if(flag == 0){
        cout<<"0";
    }else{
        cout<<fixed<<setprecision(1)<<(result/(flag*60.0));
    }
    return 0;
}