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32400sCo[sum].arrive=hh*3600+mm*60+ss-beg;//轉(zhuǎn)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里的sum寫成了i。。。。導(dǎo)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(dāng)前窗口都沒人；為什么我一開始這里定義為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(wù)。for(int tm=0;no!=n;tm++){//int tm=0;tm<32400;tm++，一開始for循環(huán)條件是這個，但是發(fā)現(xiàn)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(wù)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(dāng)前正好有結(jié)束的。//下一位顧客進來win[i]=-1;}}}for(int i=0;i<k;i++){//如果有空，那么就開始放。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;//在這里不會四舍五入！！！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(shù)組就錯了。之后還答案錯誤，原來是數(shù)組下標寫錯了。感謝牛客網(wǎng)，通不過的話會有樣例，能根據(jù)樣例去修改代碼！

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轉(zhuǎn)載于:https://www.cnblogs.com/BlueBlueSea/p/9384980.html

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