PAT 1026 Table Tennis

Posted 2021-01-03 mr-stn

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

注意点： 留意题中的条件， 每个人用桌的时间不能超过120分钟

 1 #include<iostream>
 2 #include<vector>
 3 #include<algorithm>
 4 using namespace std;
 5 struct Node{
 6   int arrivalTime, beginTime, endTime, serveTime;
 7   Node(int arrival, int serve){
 8     arrivalTime = arrival;
 9     serveTime = serve;
10   }
11 };
12 
13 bool cmp(Node& a, Node& b){return a.arrivalTime<b.arrivalTime;}
14 bool cmp1(Node& a, Node& b){return a.beginTime<b.beginTime;}
15 
16 int main(){
17   int n, i, j;
18   //v1记录普通人, v2记录会员， v记录合并后的结果
19   vector<Node> v1, v2, v;
20   scanf("%d", &n);
21   for(i=0; i<n; i++){
22     int h, m, s, t, vip;
23     scanf("%d:%d:%d %d %d", &h, &m, &s, &t, &vip);
24     //我的天啊 找了好久的bug ，居然在这里， 每个人的用桌时间不能超过120min
25     if(t>120) t = 120; 
26     if(vip==0) v1.push_back(Node(h*3600+60*m+s, t*60));
27     else v2.push_back(Node(h*3600+60*m+s, t*60));
28   }
29   sort(v1.begin(), v1.end(), cmp);
30   sort(v2.begin(), v2.end(), cmp);  
31   int k, m;
32   scanf("%d%d", &k, &m);
33   /*
34     table: 记录该桌结束的时间
35     vipTable: 记录该桌是否是vip桌
36     tableNum: 记录该桌服务的人数
37   */
38   vector<int> vipTable(k+1, false), tableNum(k+1, 0), table(k+1, 8*3600);
39   for(i=0; i<m; i++){
40     int temp;
41     scanf("%d", &temp);
42     vipTable[temp] = true;
43   }
44   int idx1=0, idx2=0;
45   for(i=0; i<n; i++){
46     int firstEnd=1, firstvip=-1;
47     for(j=1; j<=k; j++) //先找到结束时间最早的， 编号最小的桌号
48       if(table[firstEnd]>table[j]) firstEnd = j;
49     for(j=1; j<=k && firstvip==-1; j++)//找到结束时间最早的， 编号最小的， vip桌的桌号
50       if(table[firstEnd]==table[j] && vipTable[j]) firstvip = j;
51       /*
52         三种情况下 vip会员占据球桌
53         1.已经没有普通人
54         2.vip到达时间比普通人到达时间早 选择结束时间最早的  编号最小的桌子
55         3.vip到达时间比普通人到达时间晚， 但是有vip球桌空闲 选择编号最小的vip桌子
56       */
57     if(idx1>=v1.size() || (firstvip!=-1 && idx2<v2.size() && (v2[idx2].arrivalTime<=table[firstvip] )) || (idx2<v2.size() && v2[idx2].arrivalTime<v1[idx1].arrivalTime) ){
58         if(firstvip!=-1) firstEnd = firstvip;
59         
60         v2[idx2].beginTime = table[firstEnd]<v2[idx2].arrivalTime ? v2[idx2].arrivalTime : table[firstEnd];
61       table[firstEnd] = v2[idx2].beginTime + v2[idx2].serveTime;
62       v2[idx2].endTime = table[firstEnd];
63         if(v2[idx2].beginTime<21*3600){ //在21点之前得到桌子试用权的才能被记录
64             v.push_back(v2[idx2]);
65             tableNum[firstEnd]++;
66          }
67       idx2++;
68     }else{
69         v1[idx1].beginTime = table[firstEnd]<v1[idx1].arrivalTime ? v1[idx1].arrivalTime : table[firstEnd];
70       table[firstEnd]  = v1[idx1].beginTime + v1[idx1].serveTime;
71       v1[idx1].endTime = table[firstEnd];
72         if(v1[idx1].beginTime<21*3600){
73             v.push_back(v1[idx1]);
74             tableNum[firstEnd]++;
75            }
76       idx1++;
77     }
78   }
79   sort(v.begin(), v.end(), cmp1);
80   for(i=0; i<v.size(); i++)
81     printf("%02d:%02d:%02d %02d:%02d:%02d %d
", v[i].arrivalTime/3600, v[i].arrivalTime%3600/60, v[i].arrivalTime%60, v[i].beginTime/3600, v[i].beginTime%3600/60, v[i].beginTime%60, (v[i].beginTime-v[i].arrivalTime+30)/60);
82   printf("%d", tableNum[1]);
83   for(i=2; i<=k;  i++) printf(" %d", tableNum[i]);
84   return 0;
85 }