PTA-1017——Queueing at Bank（部分正确，查错半天没找到错误的

发布时间：2020-12-14 02:36:02 所属栏目：Windows 来源：网络整理

导读：题目: 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 b

题目:

Suppose a bank has?

Now given the arriving time?

Input Specification:

Each input file contains one test case. For each case,the first line contains 2 numbers:?

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

分析:

感觉代码跟其它博客思路差不多，但是就是通不过，不知道为啥。

代码:

 1 #include<iostream>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 int k,n;
 6 struct Customer{
 7     int need;    //需要的时间 
 8     int arrive;    //到达的时间 
 9     int wait;    //等待的时间 
10 }customer[10001];    //客户信息 
11 int now[101];    //窗口当前时间 
12 int win[101];    //窗口当前客户编号 
13 
14 bool cmp(Customer a,Customer b){    //根据到达时间先后快排 
15     if(a.arrive<b.arrive){
16         return true;
17     }else{
18         return false;
19     }
20 }
21 
22 int main(){
23     cin>>n>>k;
24     int num=1;    //客户数量 
25     memset(customer,0,sizeof(customer));
26     memset(win,sizeof(win));
27     for(int i=1;i<=n;i++){
28         int hour,minute,second,time;
29         scanf("%d:%d:%d %d",&hour,&minute,&second,&time);
30         if(time>60){
31             time=60;
32         }
33         int arriveTime=hour*60*60+minute*60+second;
34         if(arriveTime>17*60*60){    //如果到达时间晚于17点，忽略不计 
35             continue;
36         }else{
37             customer[num].arrive=arriveTime;
38             customer[num].need=time*60;        //统一化成秒 
39             num++;
40         }
41     }
42     num--;    //因为是从1开始计的，最终数量要减1 
43     sort(customer,customer+num,cmp);
44     for(int i=1;i<=num;i++){    //如果早于8点，起始等待时间设定为开门前的等待时间，到达时间改为八点 
45         if(customer[i].arrive<8*60*60){
46             customer[i].wait=8*60*60-customer[i].arrive;
47             customer[i].arrive=8*60*60;    
48         }
49     }
50     for(int i=1;i<=num;i++){
51         if(i<=k){        //每个窗口的第一个客户 
52             win[i]=i;
53             now[i]=customer[i].arrive;    //每个窗口的初始时间是最早进入的客户的到达时间 
54         }else{
55             int it=0;    //最早结束的窗口 
56             int finish=10000000;    //最早结束的时间 
57             for(int j=1;j<=k;j++){    //寻找最早结束的窗口 
58                 if(now[j]+customer[win[j]].need<finish){
59                     finish=now[j]+customer[win[j]].need;
60                     it=j;
61                 }
62             }
63             win[it]=i;
64             customer[i].wait+=max(finish-customer[i].arrive,0);    //等待时间最小为0，避免出现等待时间为负数的情况 
65             now[it]=max(finish,customer[i].arrive); 
66         }
67     }
68     float ans=0.0;
69     for(int i=1;i<=num;i++){
70         ans+=customer[i].wait;
71     }
72     if(num!=0){
73         ans/=num*60;    //最终结果不是秒，是分钟 
74     }
75     printf("%.1f",ans);
76     return 0;
77 }

（编辑：李大同）

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