学步园

       

这道题本身不难。但一直没有过，问题找了很久没没发现。看了别人的代码之后，才发现原来题目中讲的“Note that since the bank is closed everyday after 17:00, for those customers who cannot be served before 17:00, you must output "Sorry" instead.” 其实是说开始服务时间不能超过17:00而不是服务结束时间。
         这题方法很多，无非都是排队。可以是用循环数组，也可以直接调用queue.方法一是自己写的，完完全全的模拟。方法二是看过其他人(http://blog.csdn.net/cloudbye/article/details/7792140)的思路后修正的，更简单，也不易错。

方法一：

#include <iostream>
#include <queue>
using namespace std;

#define CUSTOMER_MAX 1000+1
#define INF 0x6fffffff  

#ifndef LOCAL
//	#define LOCAL
#endif LOCAL

int n; // number of windows <=20
int m ;// queue capacity  <=10
int k; // customers  <=1000
int q; // query times <=1000

int  ProcessTime[CUSTOMER_MAX]; // 
queue<int> que[20];
queue<int >Wait[20];
int currTime = 0;
int LeaveTime[CUSTOMER_MAX];
int Timebase[20] = {0};

int main()
{
#ifdef LOCAL
	freopen("input.txt","r",stdin);
	freopen("output.txt","w",stdout);
#endif
	cin>>n>>m>>k>>q;
	for(int i=0;i<k;i++)
	{
		cin>>ProcessTime[i];
	}
	int index;
	int top = 0;
	for(int i = 0;i<2*k;i++) 
	{
		int min_len  = m;
		if(top !=k) // if there are any customer not in line
		{
			for(int j=0;j<n;j++) 
			{
				if(min_len > que[j].size() )
				{
					min_len = que[j].size();
					index = j;
				}
			}
		}
		if(min_len != m) // find minimum queue 
		{
			que[index].push(top);
			Wait[index].push(ProcessTime[top]);
			top++;
		}else  // no queue available or no customer not in line, then customer pop 
		{
			long min_wait = INF;
			bool empty = true;
			for(int j=0;j<n;j++)
			{
				if(Wait[j].empty()) continue;
				if(min_wait > Timebase[j]+Wait[j].front())  // find current minimum wait time
				{
					min_wait = Timebase[j]+Wait[j].front();
					index = j;
					empty = false;
				}
			}
			if(empty) break;
			Timebase[index] += Wait[index].front();
			LeaveTime[que[index].front()] = Timebase[index];
			que[index].pop();
			Wait[index].pop();
		}
	}
	
	//60*9	
	int qq;
	for(int i=0;i<q;i++)
	{
		cin>>qq;
		qq--;
                //if(LeaveTime[qq]<=60*9) // 题意看错了，悲了个剧的。是服务开始时间不超过17:00，而不是结束时间
                if(LeaveTime[qq]-ProcessTime[qq]<60*9)
		{			
			int hour = LeaveTime[qq]/60;
			int second = LeaveTime[qq]%60;
			  printf("%02d:%02d\n",8+hour,second);  
		}
		else
			printf("Sorry\n");
	}
	
#ifdef  LOCAL
	system("PASUE");
#endif LOCAL

	return 0;
}

方法二：

#include <iostream>  
#include <queue>  
#include <stdio.h>  
using namespace std;  

#define CUSTOMER_MAX 1000+5
#define INF 0x6fffffff    

int n; // number of windows <=20  
int m ;// queue capacity  <=10  
int k; // customers  <=1000  
int q; // query times <=1000  

int  ProcessTime[CUSTOMER_MAX]; //   
queue<int> que[20+5];  
int LeaveTime[CUSTOMER_MAX]={0};  
int Timebase[20+5] = {0};

int main()  
{  
	cin>>n>>m>>k>>q;  
	for(int i=0;i<k;i++)  
	{  
		cin>>ProcessTime[i];  
	}

	int index;
	int top = 0;  

	// n*m
	for (int i=0;top<m*n && top<k;top++)
	{
		que[i].push(top);
		LeaveTime[top] = Timebase[i]+ProcessTime[top];
		Timebase[i] = LeaveTime[top];
		i = (i+1)%n;
	}
	for(;top<k;top++)  // pop and push
	{
		
		// find earliest leave customer
		int min_wait = INF;
		int found = false;
		for(int j=0;j<n;j++)
		{
			int cus = que[j].front();
			if(min_wait > LeaveTime[cus])  // can not be empty
			{
				min_wait = LeaveTime[cus];
				index = j;
				found = true;			
			}
		}					
		que[index].pop();  
		que[index].push(top);
		LeaveTime[top] = Timebase[index] + ProcessTime[top];  
		Timebase[index] = LeaveTime[top];
	}

	while(q--)
	{
		int qq;
		cin>>qq;  
		qq--;  
		if(LeaveTime[qq]-ProcessTime[qq]>=60*9)
		{
			printf("Sorry\n");  		
		}
		else
		{             
			int hour = LeaveTime[qq]/60;  
			int second = LeaveTime[qq]%60;  
			printf("%02d:%02d\n",8+hour,second);    
		}  
	}
	return 0;  
}