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 = 0; i < K; ++i)
    {
        scanf("%d", &cus[i].process);
        cus[i].leave = INF;
    }
    //process
    vector<queue<int>> winQueue(N);
    vector<int> timeBase(N, 0);
    int p;
    for(p = 0; 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 = -1;
        for(int j = 0; 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 = 0; i < Q; ++i)
    {
        int q;
        scanf("%d",&q);
        q--;
        if(cus[q].leave-cus[q].process >= 540)
             printf("Sorry
");
        else 
            printf("%02d:%02d
", 8+cus[q].leave/60, cus[q].leave%60);
    }
    return 0;
}