1017. Queueing at Bank (25)

题目地址：https://www.patest.cn/contests/pat-a-practise/1017

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, YueSuppose 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 awindow 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 - theprocessing 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<iostream>#include<algorithm>#include<string.h>using namespace std;struct customer{int arrive_time, server_time, long_time, wait_time;//到达时间，开始服务的时间，时长，等待时间int server_window;//在哪个窗口进行服务} cus [10000];int server_time[100];//服务窗口可以开始进行下次服务的时间，用数字表示，0代表8：00：00，//1代表8：00：01，一次类推int time2num(int hour, int minute, int second){// 8:00：00 = 0， 8：00：01 = 1int base = 8 * 60 * 60;int time = hour * 3600 + minute * 60 + second;return time - base;}bool compare(customer a, customer b){return a.arrive_time < b.arrive_time;}/*找出最早可以提供服务的窗口*/int find_server(int k){int num = -1, min = -1;for (int i = 0; i < k; i++){if (server_time[i] < min || min == -1){num = i;min = server_time[i];}}return num;}int main(){int n, k;cin >> n >> k;int count = 0;customer temp;int hour, minute, second, latest;latest = time2num(17, 0, 0);memset(server_time, 0, 100*sizeof(int));for (int i = 0; i < n; i++){cin >> hour;getchar();cin >> minute;getchar();cin >> second;getchar();temp.arrive_time = time2num(hour, minute, second);cin >> temp.long_time;temp.long_time *= 60;if (temp.arrive_time <= latest){//17：00以后来的不服务cus[count] = temp;if (cus[count].long_time > 3600)//每个人最多被服务1小时cus[count].long_time = 3600;count++;}}sort(cus, cus + count, compare);//最后能被服务的只有count个人for (int i = 0; i < count; i++){cus[i].server_window = find_server(k);cus[i].server_time = server_time[cus[i].server_window];if (cus[i].server_time < cus[i].arrive_time)//服务时间不可能早于到达时间cus[i].server_time = cus[i].arrive_time;cus[i].wait_time = cus[i].server_time - cus[i].arrive_time;server_time[cus[i].server_window] = cus[i].server_time + cus[i].long_time;}int total = 0;double average;for (int i = 0; i < count; i++){total += cus[i].wait_time;}average = total * 1.0 / 60 / count;printf("%.1f\n", average);return 0;}