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
#include<cstdio> #include<algorithm> #include<vector> using namespace std; const int maxn = 111; const int INF = 1000000000; struct Customer{ int comeTime,serveTime; }newCustomer; vector<Customer> custom; int convert(int h,int m,int s){ return h*3600+m*60+s; } bool cmp(Customer a,Customer b){ return a.comeTime < b.comeTime; } int endTime[maxn]; int main(){ int n,k,totTime = 0; scanf("%d%d",&n,&k); int stTime = convert(8,0,0); int edTime = convert(17,0,0); for(int i = 0; i < k; i++) endTime[i] = stTime; for(int i = 0; i < n; i++){ int h,m,s,serveTime; scanf("%d:%d:%d%d",&h,&m,&s,&serveTime); int comeTime = convert(h,m,s); if(comeTime > edTime) continue; newCustomer.comeTime = comeTime; newCustomer.serveTime = serveTime <= 60 ? serveTime * 60 : 3600; custom.push_back(newCustomer); } sort(custom.begin(),custom.end(),cmp); for(int i = 0; i < custom.size(); i++){ int idx = -1, minEndTime = INF; for(int j = 0 ; j < k ; j++){ if(endTime[j] < minEndTime){ idx = j; minEndTime = endTime[j]; } } if(endTime[idx] <= custom[i].comeTime){ endTime[idx] = custom[i].comeTime + custom[i].serveTime; }else{ totTime += (endTime[idx] - custom[i].comeTime); endTime[idx] += custom[i].serveTime; } } if(custom.size() == 0) printf("0.0"); else printf("%.1f",totTime/60.0/custom.size()); return 0; }
【推荐】国内首个AI IDE,深度理解中文开发场景,立即下载体验Trae
【推荐】编程新体验,更懂你的AI,立即体验豆包MarsCode编程助手
【推荐】抖音旗下AI助手豆包,你的智能百科全书,全免费不限次数
【推荐】轻量又高性能的 SSH 工具 IShell:AI 加持,快人一步
· 基于Microsoft.Extensions.AI核心库实现RAG应用
· Linux系列:如何用heaptrack跟踪.NET程序的非托管内存泄露
· 开发者必知的日志记录最佳实践
· SQL Server 2025 AI相关能力初探
· Linux系列:如何用 C#调用 C方法造成内存泄露
· 震惊!C++程序真的从main开始吗?99%的程序员都答错了
· 别再用vector<bool>了!Google高级工程师:这可能是STL最大的设计失误
· 单元测试从入门到精通
· 【硬核科普】Trae如何「偷看」你的代码?零基础破解AI编程运行原理
· 上周热点回顾(3.3-3.9)