PAT 甲级 1017. Queueing at Bank

最新推荐文章于 2022-12-01 11:01:31 发布

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本文介绍了一个银行排队系统的模拟算法，该算法通过计算不同客户的到达时间和处理时间，来预测每个客户的平均等待时间。采用C++实现，包括使用结构体存储客户信息、对客户按到达时间排序及动态分配窗口处理客户请求等步骤。

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原题传送门

题意：有N个客户，K个窗口。已知每个客户的到达时间和需要的时长（不超过60min，超过的会被压缩成60min），如果有窗口就依次过去，如果没有窗口就在黄线外等候（黄线外只有一个队伍，先来先服务），求客户的平均等待时长（十进制，单位：分钟）。银行开放时间为8点到17点，再8点之前不开门，8点之前来的人都要等待，在17点后来的人不被服务。
< algorithm >中sort()对vector进行排序
读取标准时间用scanf()
精确到秒，先全部转化为秒来计算

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

struct node {
    int come, time;
} customer;

bool cmp1(node a, node b) {
    return a.come < b.come;
}

int main(int argc, const char * argv[]) {
    int N, K;
    cin >> N >> K;
    vector<node> line;
    for(int i = 0; i < N; ++i) {
        int hour, minute, second, time;
        scanf("%d:%d:%d %d", &hour, &minute, &second, &time);
        if(time > 60)
            time = 60;
        int cometime = hour * 3600 + minute * 60 + second;
        if(cometime > (17*60*60)) // 17点
            continue;
        customer.come = cometime;
        customer.time = time * 60;
        line.push_back(customer);
    }
    sort(line.begin(), line.end(), cmp1);
    vector<int> window(K, (8*60*60)); // 存储各窗口的结束时间，大小为K，初始化为8点
    double result = 0.0;
    for(int i = 0; i < line.size(); ++i) {
        int index = 0; // 窗口号
        int minfinish = window[0]; // 各窗口中的最早结束时间

        // 循环检查各窗口，如果有窗口结束时间更早，则更新最早结束时间和窗口号：
        for(int j = 1; j < K; ++j) {
            if(minfinish > window[j]) {
                minfinish = window[j];
                index = j;
            }
        }

        // 确定下一个服务的窗口号和时间后：与客户到达时间进行比较
        if(window[index] <= line[i].come) { // 如果在客户到达之前，客户无等待
            window[index] = line[i].come + line[i].time;
        } else { // 在客户到达之后，更新等待时间result
            result += (window[index] - line[i].come);
            window[index] += line[i].time;
        }
    }
    if(line.size() == 0)
        printf("0.0");
    else
        printf("%.1f", result / 60.0 / line.size());

    return 0;
}

附原题：

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.