1026 Table Tennis (30 分)

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:

比較複雜的模擬題，
題目大意：k張桌子，球員到達後老是選擇編號最小的桌子。若是訓練時間超過2h會被壓縮成2h，若是到達時候沒有球桌空閒就變成隊列等待。
k張桌子中m張是vip桌，若是vip桌子有空閒，並且隊列裏面有vip成員，那麼等待隊列中的第一個vip球員會到最小的vip球桌訓練。
若是vip桌子空閒可是沒有vip來，那麼就分配給普通的人。若是沒有vip球桌空閒，那麼vip球員就看成普通人處理。
給出每一個球員的到達時間、要玩多久、是否是vip（是爲1不是爲0）。給出球桌數和全部vip球桌的編號，QQ全部在關門前獲得訓練的球員的到達時間、
訓練開始時間、等待時長（取整數，四捨五入），營業時間爲8點到21點。若是再21點後尚未開始玩的人，就再也不玩，不須要輸出～

個人代碼牛客經過了，pat有三組過不了，不曉得bug在哪爲啥。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

  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;//從小到大排序
 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 爲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爲非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 }

 

柳諾的代碼過了pat可是過不了牛客。。。
下面是連接
https://www.liuchuo.net/archives/2955