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：k张桌子，球员到达后总是选择编号最小的桌子。如果训练时间超过2h会被压缩成2h，如果到达时候没有球桌空闲就变成队列等待。
k张桌子中m张是vip桌，如果vip桌子有空闲，而且队列里面有vip成员，那么等待队列中的第一个vip球员会到最小的vip球桌训练。如果vip桌子空闲但是没有vip来，那么就分配给普通的人。如果没有vip球桌空闲，那么vip球员就当作普通人处理。
给出每个球员的到达时间、要玩多久、是不是vip（是为1不是为0）。给出球桌数和所有vip球桌的编号，QQ所有在关门前得到训练的球员的到达时间、训练开始时间、等待时长（取整数，四舍五入），营业时间为8点到21点。如果再21点后还没有开始玩的人，就不再玩，不需要输出～
分析：在变量t中将输入时间转化为秒的形式，使用T数组存储某个时刻到达的客户玩耍的时间，使用V数组记录某个时刻到达的客户是否是VIP，在vip数组中存储某张桌子是不是vip专属，Table记录每个桌子上的客户剩余使用时间，队列Wait，vWait记录还在排队的普通、VIP客户，now记录当前要去玩耍的客户的到达时刻，nowt记录当前要去玩耍的桌子，num数组记录每张桌子使用的人数，AnsI数组和AnsO数组分别记录客户的到达时刻和开始玩耍时刻，table与vtalbe分别记录当前可使用的最小的普通、VIP桌子。思路：将客户到达时间的时分秒统一转换为秒，同时注意每对客户最多让玩2个小时。以秒为单位，遍历从8点到21点的每个时刻，判断当前是否有新到达的普通或VIP客户，将其加入对应的队列中。同时遍历每张桌子，将有人的桌子的客户剩余游玩时刻-1，同时判断空闲的最小编号的普通、VIP桌子编号。当VIP排队队列有人时，判断是否有空的VIP桌子，有的话直接进行记录；如果没有，判断是否有空缺的普通桌子，并是否存在更早到的普通客户；没有的话，就记录当前VIP客户使用了普通桌子。同理，判断普通客户的桌子使用情况。在输出时，将秒转化为时、分、秒的格式即可~
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#include <iostream> #include <vector> #include <queue> #include <map> using namespace std; int n, m, k, H, M, S, t, table, vtable, cnt, now, nowt, T[100000], V[100000], num[10001], AnsI[10001], AnsO[10001], vip[10001]; map<int, int> Table; queue<int> Wait, vWait; int main() { scanf("%d", &n); for (int i = 0; i < n; i++) { scanf("%d:%d:%d", &H, &M, &S); t = H * 3600 + M * 60 + S; scanf("%d %d", &T[t], &V[t]); T[t] = min(T[t], 120) * 60; } scanf("%d %d", &m, &k); for (int i = 0; i < k; i++) { scanf("%d", &t); vip[t] = 1; } for (int Time = 28800; Time < 75600; Time++, table = vtable = now = 0) { if (T[Time] && V[Time]) vWait.push(Time); else if (T[Time]) Wait.push(Time); for (int i = 1; i <= m; i++) { if (Table[i] > 0) Table[i]--; if (Table[i] == 0 && vip[i] && vtable == 0) vtable = i; if (Table [i] == 0 && table == 0) table = i; } if (!vWait.empty() && (table || vtable)) { now = vWait.front(); nowt = vtable; if (vtable != 0) vWait.pop(); else { nowt = table; if (!Wait.empty() && Wait.front() < vWait.front()) { now = Wait.front(); Wait.pop(); } else vWait.pop(); } } else if (!Wait.empty() && (table || vtable)) { if (table != 0) nowt = table; else nowt = vtable; now = Wait.front(); Wait.pop(); } if (now == 0) continue; Table[nowt] = T[now]; AnsI[cnt] = now; AnsO[cnt++] = Time; num[nowt]++; } for (int i = 0; i < cnt; i++) printf("%02d:%02d:%02d %02d:%02d:%02d %d\n", AnsI[i] / 3600, AnsI[i] % 3600 / 60, AnsI[i] % 60, AnsO[i] / 3600, AnsO[i] % 3600 / 60, AnsO[i] % 60, (AnsO[i] - AnsI[i] + 30) / 60); for (int i = 1; i <= m; i++) { if (i != 1) printf(" "); printf("%d", num[i]); } return 0; } |
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