1026 Table Tennis (30 分)
标签：min test its art roc customer following ack efi
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.
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.
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.
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
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
标签：min test its art roc customer following ack efi
原文地址：https://www.cnblogs.com/zllwxm123/p/11072752.html
