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
More complex simulation title,
subject to the effect: k tables, always choose the players to reach the minimum number of tables. If the training time exceeds 2h will be compressed into 2h, arrival time if there is no table, which becomes idle waiting queue.
k is a table in the m tables vip, vip table, if there is idle, and there are queues vip member, then wait in the queue will be the first player to vip vip table, which is a minimum of training.
If the table vip vip idle but did not come, then assigned to the ordinary person. If there is no vip table, which is free, then it is treated as ordinary players vip treatment.
Each player is given arrival time, how long to play, is not the vip (is not 0 to 1). Given pool table number and the number of all the vip pool table, QQ all be trained before the closing time of arrival of the players,
a training start time, length (rounded, rounded) waiting, open from 8:00 to 21:00. If after another 21-point people have not started playing, not playing, no output -
my code through the cattle off, pat there are three groups had not, I do not know why bug where.
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 }
Liu promise code after a cow pat, but I could not pass off. . .
Here is the link
https://www.liuchuo.net/archives/2955