Pat(Advanced Level)Practice--1026(Table Tennis)

Pat1026代码

题目描写叙述：

A table tennis club has N tables available to the public. The tables are numbered from 1 to N. For any pair of players, if there are some tables open when they arrive, they will be assigned to the available table with the smallest number. If all the tables
are occupied, they will have to wait in a queue. It is assumed that every pair of players can play for at most 2 hours.

Your job is to count for everyone in queue their waiting time, and for each table the number of players it has served for the day.

One thing that makes this procedure a bit complicated is that the club reserves some tables for their VIP members. When a VIP table is open, the first VIP pair in the queue will have the priviledge to take it. However, if there is no VIP in the queue, the next
pair of players can take it. On the other hand, if when it is the turn of a VIP pair, yet no VIP table is available, they can be assigned as any ordinary players.

Input Specification:

Each input file contains one test case. For each case, the first line contains an integer N (<=10000) - the total number of pairs of players. Then N lines follow, each contains 2 times and a VIP tag: HH:MM:SS - the arriving time, P - the playing time in minutes
of a pair of players, and tag - which is 1 if they hold a VIP card, or 0 if not. It is guaranteed that the arriving time is between 08:00:00 and 21:00:00 while the club is open. It is assumed that no two customers arrives at the same time. Following the players'
info, there are 2 positive integers: K (<=100) - the number of tables, and M (< K) - the number of VIP tables. The last line contains M table numbers.

Output Specification:

For each test case, first print the arriving time, serving time and the waiting time for each pair of players in the format shown by the sample. Then print in a line the number of players served by each table. Notice that the output must be listed in chronological
order of the serving time. The waiting time must be rounded up to an integer minute(s). If one cannot get a table before the closing time, their information must NOT be printed.

Sample Input:

9
20:52:00 10 0
08:00:00 20 0
08:02:00 30 0
20:51:00 10 0
08:10:00 5 0
08:12:00 10 1
20:50:00 10 0
08:01:30 15 1
20:53:00 10 1
3 1
2

Sample Output:

08:00:00 08:00:00 0
08:01:30 08:01:30 0
08:02:00 08:02:00 0
08:12:00 08:16:30 5
08:10:00 08:20:00 10
20:50:00 20:50:00 0
20:51:00 20:51:00 0
20:52:00 20:52:00 0
3 3 2

这个排序模拟题。真是延续了PAT坑人的一贯作风啊；

參考了网上的一些代码，下面几点须要注意：

1.当有多个乒乓球台空暇时，vip顾客到了会使用最小id的vip球台，而不是最小id的球台，測试下面用例：

2
10:00:00 30 1
12:00:00 30 1
5 1
3

输出正确结果应为：

10:00:00 10:00:00 0

12:00:00 12:00:00 0

0 0 2 0 0

2.题目要求每对顾客玩的时间不超过2小时。那么当顾客要求玩的时间>2小时的时候，应该截断控制。測试下面用例：

2
18:00:00 180 1
20:00:00 60 1
1 1
1

输出的正确结果应为：

18:00:00 18:00:00 0

20:00:00 20:00:00 0

2

3.尽管题目中保证客户到达时间在08:00:00到21:00:00之间。可是依据最后的8个case来看，里面还是有不在这个时间区间内到达的顾客，所以建议还是稍加控制。測试下面用例：

1
21:00:00 80 1
1 1
1

输出的正确结果应为：

0

4.题目中说的round up to an integer minutes是严格的四舍五入。须要例如以下做：

wtime = (stime - atime + 30) / 60

而不是：

wtime = (stime - atime + 59) / 60

     AC代码：

#include<cstdio>
#include<vector>
#include<algorithm>
 
using namespace std;
 
class Player
{
	public:
		int arrive;//arrive time
		int vip;//vip flag
		int playtime;//the time player play
		bool operator<(const Player r)const
		{
			return arrive<r.arrive;
		}
};
 
class Table
{
	public:
		int freetime;//the time table can be used
		int vip;//vip falg
		int ID;//the number of table
		int num;//the number of players the table serve
		bool operator<(const Table r)const
		{
			if(freetime!=r.freetime)
				return freetime<r.freetime;
			else
				return ID<r.ID;
		}
};
 
bool cmp(Table l,Table r)
{
	return l.ID<r.ID;
}
 
int main(int argc,char *argv[])
{
	Table t[105];
	vector<Player> p;
	vector<Player> waiting;
	int n,i,j;
	int k,m;
	scanf("%d",&n);
	for(i=0;i<n;i++)
	{
		int h,m,s;
		Player temp;
		scanf("%d:%d:%d %d %d",&h,&m,&s,&temp.playtime,&temp.vip);
		temp.arrive=h*3600+m*60+s;
		if(temp.playtime>120)
			temp.playtime=120;
		temp.playtime*=60;
		if(temp.arrive>=21*60*60||temp.arrive<8*60*60)
			continue;
		p.push_back(temp);
	}
	scanf("%d%d",&k,&m);
	for(i=1;i<=k;i++)
	{
		t[i].ID=i;
		t[i].vip=0;
		t[i].num=0;
		t[i].freetime=0;
	}
	for(i=0;i<m;i++)
	{
		int index;
		scanf("%d",&index);
		t[index].vip=1;
	}
	int timer=0;
	int cur=0;
	sort(p.begin(),p.end());
	sort(t+1,t+1+k);
	while(timer<21*60*60)
	{
		for(;cur<p.size();cur++)
		{
			if(p[cur].arrive<=timer)//假设此时没有空暇球桌，而且在timer
				waiting.push_back(p[cur]);//之前到达，player需在队列里等待
			else
				break;
		}
		if(!waiting.size())//假设等待对列为空。即此时有剩余球桌
		{
			if(cur<p.size())
			{
				timer=p[cur].arrive;
				for(i=1;i<=k;i++)//把无人使用球桌的freetime更新为timer
				{
					if(t[i].freetime<=timer)
						t[i].freetime=timer;
				}
				waiting.push_back(p[cur++]);
			}
			else
				break;
		}
		vector<Player>::iterator it;//find the first vip player in the queue
		for(it=waiting.begin();it!=waiting.end();it++)
			if(it->vip)
				break;
		int vipplayer=0;
		if(it!=waiting.end())
			vipplayer=1;
		int viptable=-1;//find the first vacant vip table
		for(i=1;i<=k&&t[i].freetime==timer;i++)
		{
			if(t[i].vip)
			{
				viptable=i;
				break;
			}
		}
		if(viptable>=1&&vipplayer)//队列中有vip客户，且有空暇vip桌子
		{
			int arrive=it->arrive;
			printf("%02d:%02d:%02d %02d:%02d:%02d %d\n",arrive/3600,
		(arrive%3600)/60,arrive%60,timer/3600,(timer%3600)/60,timer%60,
		         (timer-it->arrive+30)/60);
			t[viptable].freetime=timer+it->playtime;
			t[viptable].num++;
			waiting.erase(it);
		}
		else
		{
			int arrive=waiting[0].arrive;
			 printf("%02d:%02d:%02d %02d:%02d:%02d %d\n",arrive/3600,
			 (arrive%3600)/60,arrive%60,timer/3600,(timer%3600)/60,timer%60,
	                 (timer-arrive+30)/60);
			 t[1].freetime=timer+waiting[0].playtime;
			 t[1].num++;
			 waiting.erase(waiting.begin());
		}
		sort(t+1,t+1+k);//对桌子的空暇时间进行排序，否则对ID进行排序
		timer=t[1].freetime;//则每次都是第一个桌子先被使用
	}
	sort(t+1,t+1+k,cmp);//恢复桌子的序号
	printf("%d",t[1].num);
	for(i=2;i<=k;i++)
		printf(" %d",t[i].num);
	printf("\n");
 
	return 0;
}