1026 Table Tennis (30分)--PAT甲级真题（测试点过不了来看看排坑）

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

题目大意：给出顾客到达的时间，以及打乒乓球的时间，并且该顾客是否是VIP；并且给出一些桌子，有一些桌子是VIP桌；要去求每个顾客得到服务的时间，等待的时间，还有每个桌子服务顾客的个数；
分析：现给大家排几个坑吧
1. 顾客最多可以打2小时,输入时间为2小时以上的,改为2小时
2. VIP客户会优先选择空闲VIP桌,哪怕有编号更小的空闲非VIP桌
3. 最后等待时间需要(rounded up)四舍五入

	#include
	#include
	#include
	#include
	using namespace std; 
	const int OPEN = 3600 * 8;
	const int CLOSE = 3600 * 21;
	struct cus{
		int arvT, serT = -1, costT, flag; 
	};  
	vector<cus> Vu; //顾客
	int FreeT[101]; //桌子空闲的时间
	set<int> VipT;//vip桌子
	int SerCnt[101];//每个桌子服务的人数
	int N, K, M; 
	bool cmp(cus A, cus B) {return A.arvT < B.arvT;} 
	bool cmp2(cus A, cus B) {return A.serT < B.serT;} 
	int selectT(int arvT){
		int ret, min = 123123123;
		for (int i = 1; i <= K; i++){
			if (arvT >= FreeT[i]){
				ret = i;
				break;
			}
			else if (FreeT[i] < min){
				min = FreeT[i];
				ret = i;
			}
		}
		return ret;
	}  
	int VipInQue(int s, int time){
		int ret = -1;
		for (int i = s; i < Vu.size(); i++){
			if (Vu[i].flag == 1 && Vu[i].serT == -1 && Vu[i].arvT < time){
				ret = i;
				break;
			}
		}
		return ret;
	} 
	int vipTable(int arvT){
		int ret = -1;
		for (auto it = VipT.begin(); it != VipT.end(); it++) {
			if (FreeT[*it] <= arvT) {
				ret = *it;
				break;
			}
		} 
		return ret;
	} 
	int main(){
		scanf("%d", &N);
		int x;
		for (int i = 0; i < N; i++){
			int h, m, s;
			cus tmp;
			scanf("%d:%d:%d %d %d", &h, &m, &s, &tmp.costT, &tmp.flag);
			tmp.arvT = 3600 * h + 60 * m + s; 
			if (tmp.costT > 120)
				tmp.costT = 120;
			tmp.costT *= 60;
			Vu.push_back(tmp);
		}
		scanf("%d%d", &K, &M); 
		for (int i = 0; i < M; i++) {
			scanf("%d", &x);
			VipT.insert(x);
		}  
		fill(FreeT, FreeT + 101, OPEN);
		sort(Vu.begin(), Vu.end(), cmp);
		for (int i = 0; i < Vu.size();){
			if (Vu[i].serT != -1){//如果该顾客已经安排过，安排下一位
				i++;
				continue;
			} 
			int t = selectT(Vu[i].arvT);
			if (VipT.find(t)!= VipT.end() && Vu[i].flag == 0 && VipInQue(i + 1, FreeT[t]) != -1){
				//选中的空闲桌是vip桌，队首不是vip,并且队列里有排队的vip
				x = VipInQue(i + 1, FreeT[t]);//将这个空闲vip桌分配给队列中第一个vip
				Vu[x].serT = FreeT[t];
				FreeT[t] = Vu[x].serT + Vu[x].costT;
				if (Vu[x].serT < CLOSE)
					SerCnt[t]++;
			}
			else if (VipT.find(t) == VipT.end() && Vu[i].flag == 1 && vipTable(Vu[i].arvT) != -1) {
				//选中的桌子不是vip桌子，但是队首是vip，且有空闲的vip桌子
				x = vipTable(Vu[i].arvT); //找一个vip桌分配给队首
				Vu[i].serT = Vu[i].arvT;
				FreeT[x] = Vu[i].serT + Vu[i].costT;
				if (Vu[i].serT < CLOSE)
					SerCnt[x]++;
				i++;
			}
			else{
				if (Vu[i].arvT >= FreeT[t]) 
					Vu[i].serT = Vu[i].arvT; 
				else
					Vu[i].serT = FreeT[t]; 
				FreeT[t] = Vu[i].serT + Vu[i].costT; 
				if (Vu[i].serT < CLOSE)
					SerCnt[t]++;
				i++;
			}
		} 
		sort(Vu.begin(), Vu.end(), cmp2);
		for (int i = 0; i < Vu.size(); i++){
			if (Vu[i].serT < CLOSE)
				printf("%02d:%02d:%02d %02d:%02d:%02d %d\n", Vu[i].arvT / 3600, (Vu[i].arvT % 3600) / 60, Vu[i].arvT % 60,
					Vu[i].serT / 3600, (Vu[i].serT % 3600) / 60, Vu[i].serT % 60, (Vu[i].serT - Vu[i].arvT + 30) / 60);
		} 
		for (int i = 1; i <= K; i++){
			if (i != 1)
				printf(" ");
			printf("%d", SerCnt[i]);
		} 
		return 0;
	}