PAT A 1014. Waiting in Line (30)【队列模拟】

题目:https://www.patest.cn/contests/pat-a-practise/1014

思路：

直接模拟类的题。

线内的各个窗口各为一个队，线外的为一个，按时间模拟出队、入队。

注意点：即使到关门时间，已经在服务中的客户（窗口第一个，接待时间早于关门时间）还是可以被服务的。其它的则不服务。

 1 #include<iostream>
 2 #include<vector>
 3 #include<set>
 4 #include<map>
 5 #include<queue>
 6 #include<algorithm>
 7 #include<string>
 8 #include<string.h>
 9 using namespace std;
10 
11 int N;// (<=20, number of windows)
12 int M;// (<=10, the maximum capacity of each line inside the yellow line)
13 int K;// (<=1000, number of customers)
14 int Q;// (<=1000, number of customer queries)
15 #define INF 0x6FFFFFFF
16 typedef struct Customer
17 {
18     int process;
19     int leave;
20 }Customer;
21 
22 int main()
23 {
24     //input
25     scanf("%d%d%d%d",&N,&M,&K,&Q);
26     vector<Customer> cus(K);
27     for(int i = 0; i < K; ++i)
28     {
29         scanf("%d", &cus[i].process);
30         cus[i].leave = INF;
31     }
32     //process
33     vector<queue<int>> winQueue(N);
34     vector<int> timeBase(N, 0);
35     int p;
36     for(p = 0; p < N*M && p < K; ++p)
37     {
38         cus[p].leave = timeBase[p%N]+cus[p].process;
39         timeBase[p%N] = cus[p].leave;
40         winQueue[p%N].push(p);
41     }
42     //for somebody out of the normal queue
43     for(; p < K; ++p)
44     {
45         int mmin = INF;
46         int index = -1;
47         for(int j = 0; j < N; ++j)
48         {
49             int top = winQueue[j].front();
50             if(mmin > cus[top].leave)
51             {
52                 index = j;
53                 mmin = cus[top].leave;
54             }
55         }
56         //then pop 
57         cus[p].leave = timeBase[index]+cus[p].process;
58         timeBase[index] = cus[p].leave;
59         winQueue[index].pop();
60         winQueue[index].push(p);
61     }
62     
63     //query 
64     for(int i = 0; i < Q; ++i)
65     {
66         int q;
67         scanf("%d",&q);
68         q--;
69         if(cus[q].leave-cus[q].process >= 540)
70              printf("Sorry
");
71         else 
72             printf("%02d:%02d
", 8+cus[q].leave/60, cus[q].leave%60);
73     }
74     return 0;
75 }