1017. Queueing at Bank (25)——PAT (Advanced Level) Practise

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最新推荐文章于 2025-01-22 23:38:46 发布

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分类专栏：浙大PAT 浙大Pat 文章标签： pat 1017 Advanced

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本文详细介绍了如何通过编程解决银行排队问题，包括将时间转化为秒进行处理，剔除非有效时间段的客户，以及按秒模拟整个过程来计算所有客户的平均等待时间。通过实际案例演示了解题思路并提供了相应的AC代码。

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题目信息

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

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

解题思路

把时间转化成秒，超出17:00的时间剔除，然后按秒模拟

AC代码

#include <cstdio>
#include <algorithm>
using namespace std;

struct node{
    long long a, w;
    bool operator<(const node& b)const{
        return a < b.a;
    }
}s[10005];
long long k[105];
long long n, m, a, b, c, t, id, wait, waitp;
int main()
{
    scanf("%lld%lld", &n, &m);
    for (int i = 0; i < n; ++i){
        scanf("%lld:%lld:%lld %lld", &a, &b, &c, &s[id].w);
        s[id].w *= 60;
        s[id].a = a*60*60+b*60+c;
        if (s[id].a >= 61200){
            --id;
        }
        ++id;
    }
    sort(s, s + id);
    c = 28800; 
    while (t < id){
        bool flag = true;
        for (int i = 0; i < m; ++i){
            if (k[i])  --k[i];
        }
        while (flag && t < id && s[t].a <= c){
            flag = false;
            for (int i = 0; i < m; ++i){
                if (!k[i]) {
                    flag = true;
                    a = i;
                }
            }
            if (flag){
                k[a] = s[t].w;
                wait += c - s[t].a;
                ++t;
            }
        }
        ++c;
    }
    printf("%.1f\n", 1.0*wait/60/id);
    return 0;
}