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。

 #include<iostream>
 #include<vector>
 #include<set>
 #include<map>
 #include<queue>
 #include<algorithm>
 #include<string>
 #include<string.h>
 using namespace std;

 int N;// (<=20, number of windows)
 int M;// (<=10, the maximum capacity of each line inside the yellow line)
 int K;// (<=1000, number of customers)
 int Q;// (<=1000, number of customer queries)
 #define INF 0x6FFFFFFF
 typedef struct Customer
 {
     int process;
     int leave;
 }Customer;

 int main()
 {
     //input
     scanf("%d%d%d%d",&N,&M,&K,&Q);
     vector<Customer> cus(K);
     for(int i = ; i < K; ++i)
     {
         scanf("%d", &cus[i].process);
         cus[i].leave = INF;
     }
     //process
     vector<queue<int>> winQueue(N);
     vector<int> timeBase(N, );
     int p;
     for(p = ; p < N*M && p < K; ++p)
     {
         cus[p].leave = timeBase[p%N]+cus[p].process;
         timeBase[p%N] = cus[p].leave;
         winQueue[p%N].push(p);
     }
     //for somebody out of the normal queue
     for(; p < K; ++p)
     {
         int mmin = INF;
         int index = -;
         for(int j = ; j < N; ++j)
         {
             int top = winQueue[j].front();
             if(mmin > cus[top].leave)
             {
                 index = j;
                 mmin = cus[top].leave;
             }
         }
         //then pop
         cus[p].leave = timeBase[index]+cus[p].process;
         timeBase[index] = cus[p].leave;
         winQueue[index].pop();
         winQueue[index].push(p);
     }

     //query
     for(int i = ; i < Q; ++i)
     {
         int q;
         scanf("%d",&q);
         q--;
         if(cus[q].leave-cus[q].process >= )
              printf("Sorry\n");
         else
             printf("%02d:%02d\n", +cus[q].leave/, cus[q].leave%);
     }
     return ;
 }