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
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～
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摘自CSDN博主「柳婼」的原创文章
原文链接：https://blog.csdn.net/liuchuo/article/details/54576965
测试点分析
一开始错了四个点：测试点3、测试点5、测试点7、测试点8
测试点3：只有一个人来并关门了
// 测试点3样例
1
21:00:00 10 1
3 0
测试点5、7：都是vip用户要选择第一张空闲的vip桌，注意时间可以取等，即刚好打完到了
测试点8：最后计算等待时间的时候>=30向上进1，用(start-arrive+30)/60就行了
AC代码
#include<bits/stdc++.h>
using namespace std;
struct cus{
int arri, play, vip;
int start;
};
bool cmp1(cus a, cus b){
return a.arri < b.arri;
}
bool cmp2(cus a, cus b){
return a.start<b.start;
}
void prt(int x){
int h, m, s;
h = x/3600;
m = x/60 -h*60;
s = x%60;
printf("%02d:%02d:%02d ", h, m, s);
}
int main(){
int n, m, k;
scanf("%d", &n);
vector<cus> data(n);
for(int i=0; i<n; i++){
int h, m, s;
scanf("%d:%d:%d %d %d", &h, &m, &s, &data[i].play, &data[i].vip);
data[i].play = min(120, data[i].play)*60; //测试点4
data[i].arri = h*3600 + m*60 +s;
}
scanf("%d %d", &m, &k);
vector<int> times(m, 8*3600);
vector<int> cnt(m);
vector<bool> is_vip(m, false);
for(int i=0; i<k; i++){
int t;
scanf("%d", &t);
is_vip[t-1] = true;
}
sort(data.begin(), data.end(), cmp1);
for(int i=0; i<n; i++){
if(data[i].start) continue;
int mint=21*3600, table=-1, player=i;
for(int j=0; j<m; j++){
if(times[j]<mint){
table = j;
mint = times[j];
}
}
if(table==-1) data[i].start = mint;
else{
if(is_vip[table]&&!data[i].vip){
for(int j=i; j<n&&data[j].arri<=mint&&player==i; j++) if(data[j].vip&&!data[j].start) player=j;
if(player!=i) i--;
}
else if(data[i].vip&&!is_vip[table]){
for(int j=0; j<m; j++){ //测试点5在vip用户用第一张空闲的vip桌
if(times[j]<=data[i].arri&&is_vip[j]){ //测试点7在这个等于号
table = j; break;
}
}
}
data[player].start = max(mint, data[player].arri);
times[table] = data[player].start + data[player].play;
if(data[player].start<21*3600) cnt[table]++; //测试点3在这个判断，第一个用户就超过21点不能计数
}
}
sort(data.begin(), data.end(), cmp2);
for(int i=0; i<n; i++){
if(data[i].start>=21*3600) continue;
prt(data[i].arri);
prt(data[i].start);
printf("%d\n", (data[i].start-data[i].arri+30)/60); //测试点8在这个分钟的四舍五入
}
for(int i=0; i<m; i++) printf("%d%s", cnt[i], i==m-1?"\n":" ");
return 0;
}
