1026 Table Tennis (30 分)

Posted 2022-02-03 zllwxm123

1026 Table Tennis (30 分)

 

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

比较复杂的模拟题，
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～

我的代码牛客通过了，pat有三组过不了，不晓得bug在哪为啥。

  1 #include<bits/stdc++.h>
  2 #define _start 28800
  3 #define _end 75600
  4 #define inf 0x3f3f3f3f
  5 using namespace std;
  6 struct Node    
  7     int total, need;
  8     friend bool operator < (const Node &a, const Node &b)
  9         return a.total > b.total;
 10     
 11 node;
 12 struct Edge
 13     int endtime;
 14     int id;
 15     friend bool operator < (const Edge &a, const Edge &b)
 16         return a.endtime > b.endtime;//从小到大排序
 17     
 18 edge;
 19 priority_queue<Node> q1,q0;
 20 priority_queue<Edge> e0;
 21 priority_queue<int, vector<int>, greater<int> > free_r,vip_r;
 22 int n,x,y;
 23 int an[105];
 24 int vis[105];
 25 int main()
 26     scanf("%d", &n);
 27     int h,f,m,val,k;
 28     for(int i = 0; i < n; i++)
 29         scanf("%d:%d:%d",&h,&f,&m);
 30         scanf("%d %d",&val, &k);
 31         node.total = h*60*60+f*60+m;
 32         if(val <= 120)
 33             node.need = val*60;
 34         else
 35             node.need = 120*60;
 36         if(k)
 37             q1.push(node);
 38         else
 39             q0.push(node);
 40     
 41     scanf("%d%d",&x,&y);
 42     int p;
 43     for(int i = 0; i < y; i++)
 44         scanf("%d", &p);
 45         an[p] = 1;
 46     
 47     for(int i = 1; i <= x; i++)
 48         if(an[i])
 49             vip_r.push(i);
 50         else
 51             free_r.push(i);
 52     
 53     for(int time = _start; time < _end; time++)
 54         while(!e0.empty()&&e0.top().endtime <= time)
 55             if(an[e0.top().id])
 56                 vip_r.push(e0.top().id);
 57             else
 58                 free_r.push(e0.top().id);
 59             
 60             e0.pop();
 61         
 62         while((!q1.empty()&&q1.top().total <= time) || (!q0.empty()&&q0.top().total <= time))
 63             int a = inf,b = inf;
 64             if((!q1.empty()&&q1.top().total <= time))
 65                 a = q1.top().total;
 66             
 67             if((!q0.empty()&&q0.top().total <= time))
 68                 b = q0.top().total;
 69             
 70             if(a != inf) //a 为VIP
 71                 int id = 0;
 72                 if(!vip_r.empty())
 73                     id = vip_r.top();
 74                     vip_r.pop();
 75                 else if(!free_r.empty())
 76                     id = free_r.top();
 77                     free_r.pop();
 78                 
 79                 if(id)
 80                     int ans = time + q1.top().need;
 81                     vis[id] ++;
 82                     int miniute = (time - a)/60 + ((time - a)%60 >= 30?1:0);
 83                     printf("%02d:%02d:%02d %02d:%02d:%02d %d\n", a/3600, (a%3600)/60, a%60, time/3600, (time%3600)/60, time%60, miniute);
 84                     e0.push(ans, id);
 85                     q1.pop();
 86                 else
 87                     break;
 88                 
 89                 
 90             else    //b为非VIP
 91                 int id = 0;
 92                 if(!free_r.empty() || !vip_r.empty())
 93                     int a1 = inf, a2 = inf;
 94                     if(!free_r.empty())
 95                         a1 = free_r.top();
 96                     if(!vip_r.empty())
 97                         a2 = vip_r.top();
 98                     if(a1 < a2)
 99                         id = a1;
100                         free_r.pop();
101                     else
102                         id = a2;
103                         vip_r.pop();
104                     
105                 
106                 if(id)
107                     int ans = time + q0.top().need;
108                     vis[id] ++;
109                     int miniute = (time - b)/60 + ((time - b)%60 >= 30?1:0);
110                     printf("%02d:%02d:%02d %02d:%02d:%02d %d\n", b/3600, (b%3600)/60, b%60, time/3600, (time%3600)/60, time%60, miniute);
111                     e0.push(ans, id);
112                     q0.pop();
113                 else
114                     break;
115                 
116             
117         
118     
119     for(int i = 1; i <= x; i++)
120         printf("%d%c", vis[i], i==x?‘\n‘:‘ ‘);
121     return 0;
122 

 

柳诺的代码过了pat但是过不了牛客。。。
下面是链接
https://www.liuchuo.net/archives/2955