1017 Queueing at Bank (25)（25 分）提问
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）提问
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:
8.2ＡＣ:
#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=;
scanf("%d%d",&n,&k);
co * Co=new co[];
if(>=n||>=k){
printf("0.0");
return ;
}
int beg=*;
for(int i=;i<n;i++){
scanf("%d:%d:%d %d",&hh,&mm,&ss,&ser);
//到达是不同时到达的，
//共32400s
Co[sum].arrive=hh*+mm*+ss-beg;//转换成秒
Co[sum].serve=ser*;
if(Co[sum].arrive<)sum++;//sum表示一共有多少人。
//我的妈呀，if里的sum写成了i。。。。导致代码通不过，醉了。
}
n=sum;
sort(Co,Co+n,cmp);//按到达时间排序。
int win[k];//表示当前窗口都没人；为什么我一开始这里定义为3，一定是气懵了。。。。
fill(win,win+k,-); int no=;//还剩下多少人需要服务。
for(int tm=;no!=n;tm++){
//int tm=0;tm<32400;tm++，一开始for循环条件是这个，但是发现了问题，
//如果这样的话，就不能保证所有在17：00之前到的顾客都能服务了。
if(no==n)break;
for(int i=;i<k;i++){
if(win[i]!=-){
if(Co[win[i]].begn+Co[win[i]].serve==tm){//当前正好有结束的。
//下一位顾客进来
win[i]=-;
}
}
}
for(int i=;i<k;i++){//如果有空，那么就开始放。
if(win[i]==-&&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=;
int miu=;
for(int i=;i<n;i++){
total+=(Co[i].begn-Co[i].arrive);
// if(total%60==0){
// miu+=total/60;
// total=0;
// }
}
//miu+=1.0*total/60;//在这里不会四舍五入！！！
printf("%.1f",total/60.0/n);
return ;
}
/**
2 2
8:00:05 1
12:00:00 1 **/
//对我自己醉了，通不过就是因为瞎。。心瞎。。遇到了段错误，原来是自己一开始定义数组就错了。之后还答案错误，原来是数组下标写错了。感谢牛客网，通不过的话会有样例，能根据样例去修改代码！