1017. Queueing at Bank (25) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue 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 custo…

1017. Queueing at Bank (25) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue 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 custo…

1017 Queueing at Bank (25 分)   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…

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 ther…

题目如下: 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 an…

problem 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…

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 ther…

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 ther…

简单模拟. #include<iostream> #include<cstring> #include<cmath> #include<algorithm> #include<cstdio> #include<map> #include<queue> using namespace std; struct X { long long hh,mm,ss; long long len; long long st,ml; }s[…

有n个客户和k个窗口,给出n个客户的到达时间和需要的时长有空闲的窗口就去办理,没有的话就需要等待,求客户的平均时长.如果在8点前来的,就需要等到8点.如果17点以后来的,则不会被服务,无需考虑. 按客户的到达时间排序建立一个优先级队列,一开始放入k个窗口,初始结束时间为8*3600然后for循环客户,每次从优先级队列中取出最早结束时间的窗口如果客户比结束时间来的早,就需要等待如果客户比结束时间来的晚,就无需等待最后只要统计那些到达时间在17*3600之前的客户即可. #include <iost…