1014. Waiting in Line (30)

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本文介绍了一个排队系统的模拟程序，使用C++实现。该程序通过输入窗口数量、每个队伍的最大容量及顾客数量等参数，模拟了顾客在不同窗口前排队并被服务的过程。程序还实现了查询功能，可以查询特定顾客的等待时间。

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考察排队

#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\n");
		else 
			printf("%02d:%02d\n", 8+cus[q].leave/60, cus[q].leave%60);
	}
	return 0;
}