1017 Queueing at Bank (25)（25 分）提問

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

(25)（25 分）提問

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

ＡＣ:

#include<iostream>
#include<stdio.h>
#include<algorithm>
using  namespace std;
struct co{
    int  arrive;
    int begn;
    int serve;
    int wait;
};

bool cmp( co a , co b){
    return a.arrive<b.arrive;
}

int main()
{
    int n,k;
    int hh,mm,ss,ser,sum=0;
    scanf("%d%d",&n,&k);
    co * Co=new co[10000];
    if(0>=n||0>=k){
        printf("0.0");
        return 0;
    }
    int beg=8*3600;
    for(int i=0;i<n;i++){
        scanf("%d:%d:%d %d",&hh,&mm,&ss,&ser);
        //到達是不同時到達的，
        //共32400s
        Co[sum].arrive=hh*3600+mm*60+ss-beg;//轉換成秒
        Co[sum].serve=ser*60;
        if(Co[sum].arrive<32400)sum++;//sum表示一共有多少人。
        //我的媽呀，if裏的sum寫成了i。。。。導致代碼通不過，醉了。
    }
    n=sum;
    sort(Co,Co+n,cmp);//按到達時間排序。
    int win[k];//表示當前窗口都沒人；為什麽我一開始這裏定義為3，一定是氣懵了。。。。
    fill(win,win+k,-1);

    int no=0;//還剩下多少人需要服務。
    for(int tm=0;no!=n;tm++){
            //int tm=0;tm<32400;tm++，一開始for循環條件是這個，但是發現了問題，
            //如果這樣的話，就不能保證所有在17：00之前到的顧客都能服務了。
        if(no==n)break;
        for(int i=0;i<k;i++){
            if(win[i]!=-1){
                if(Co[win[i]].begn+Co[win[i]].serve==tm){//當前正好有結束的。
                    //下一位顧客進來
                    win[i]=-1;
                }
            }
        }
        for(int i=0;i<k;i++){//如果有空，那麽就開始放。
            if(win[i]==-1&&Co[no].arrive<=tm){
                Co[no].begn=tm;
                win[i]=no;
                no++;
                if(no==n)break;
            }
        }
    }
//    for(int i=0;i<n;i++){
//        printf("\n%d %d %d\n",Co[i].arrive,Co[i].begn,Co[i].serve);
//    }
    long long  total=0;
    int miu=0;
    for(int i=0;i<n;i++){
            total+=(Co[i].begn-Co[i].arrive);
//            if(total%60==0){
//                miu+=total/60;
//                total=0;
//        }
    }
    //miu+=1.0*total/60;//在這裏不會四舍五入！！！
    printf("%.1f",total/60.0/n);
    return 0;
}
/**
2 2
8:00:05 1
12:00:00 1

**/

//對我自己醉了，通不過就是因為瞎。。心瞎。。遇到了段錯誤，原來是自己一開始定義數組就錯了。之後還答案錯誤，原來是數組下標寫錯了。感謝牛客網，通不過的話會有樣例，能根據樣例去修改代碼！

PAT 1017 Queueing at Bank[一般]