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
 #include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
using namespace std;
struct person {
int arrive, start, time;
bool vip;
}tempperson;
struct tablenode {
int end = * , num;
bool vip;
};
bool cmp1(person a, person b) {
return a.arrive < b.arrive;
}
bool cmp2(person a, person b) {
return a.start < b.start;
}
vector<person> player;
vector<tablenode> table;
void alloctable(int personid, int tableid) {
if (player[personid].arrive <= table[tableid].end)
player[personid].start = table[tableid].end;
else
player[personid].start = player[personid].arrive;
table[tableid].end = player[personid].start + player[personid].time;
table[tableid].num++;
}
int findnextvip(int vipid) {
vipid++;
while (vipid < player.size() && player[vipid].vip == false) vipid++;
return vipid;
}
int main() {
int n, k, m, viptable;
scanf("%d", &n);
for (int i = ; i < n; i++) {
int h, m, s, temptime, flag;
scanf("%d:%d:%d %d %d", &h, &m, &s, &temptime, &flag);
tempperson.arrive = h * + m * + s;
tempperson.start = * ;
if (tempperson.arrive >= * ) continue;
tempperson.time = temptime <= ? temptime * : ;
tempperson.vip = ((flag == ) ? true : false);
player.push_back(tempperson);
}
scanf("%d%d", &k, &m);
table.resize(k + );
for (int i = ; i < m; i++) {
scanf("%d", &viptable);
table[viptable].vip = true;
}
sort(player.begin(), player.end(), cmp1);
int i = , vipid = -;
vipid = findnextvip(vipid);
while (i < player.size()) {
int index = -, minendtime = ;
for (int j = ; j <= k; j++) {
if (table[j].end < minendtime) {
minendtime = table[j].end;
index = j;
}
}
if (table[index].end >= * ) break;
if (player[i].vip == true && i < vipid) {
i++;
continue;
}
if (table[index].vip == true) {
if (player[i].vip == true) {
alloctable(i, index);
if (vipid == i) vipid = findnextvip(vipid);
i++;
}
else {
if (vipid < player.size() && player[vipid].arrive <= table[index].end) {
alloctable(vipid, index);
vipid = findnextvip(vipid);
}
else {
alloctable(i, index);
i++;
}
}
}
else {
if (player[i].vip == false) {
alloctable(i, index);
i++;
}
else {
int vipindex = -, minvipendtime = ;
for (int j = ; j <= k; j++) {
if (table[j].vip == true && table[j].end < minvipendtime) {
minvipendtime = table[j].end;
vipindex = j;
}
}
if (vipindex != - && player[i].arrive >= table[vipindex].end) {
alloctable(i, vipindex);
if (vipid == i) vipid = findnextvip(vipid);
i++;
}
else {
alloctable(i, index);
if (vipid == i) vipid = findnextvip(vipid);
i++;
}
}
}
}
sort(player.begin(), player.end(), cmp2);
for (i = ; i < player.size() && player[i].start < * ; i++) {
printf("%02d:%02d:%02d ", player[i].arrive / , player[i].arrive % / , player[i].arrive % );
printf("%02d:%02d:%02d ", player[i].start / , player[i].start % / , player[i].start % );
printf("%.0f\n", round((player[i].start - player[i].arrive) / 60.0));
}
for (int i = ; i <= k; i++) {
if (i != ) printf(" ");
printf("%d", table[i].num);
}
return ;
}

PAT甲级——A1026 Table Tennis的更多相关文章

  1. PAT甲级1026. Table Tennis

    PAT甲级1026. Table Tennis 题意: 乒乓球俱乐部有N张桌子供公众使用.表的编号从1到N.对于任何一对玩家,如果有一些表在到达时打开,它们将被分配给具有最小数字的可用表.如果所有的表 ...

  2. PAT 甲级 1026 Table Tennis(模拟)

    1026. Table Tennis (30) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue A table ...

  3. PAT 甲级 1026 Table Tennis (30 分)(坑点很多,逻辑较复杂,做了1天)

    1026 Table Tennis (30 分)   A table tennis club has N tables available to the public. The tables are ...

  4. PAT甲级1026 Table Tennis【模拟好题】

    题目:https://pintia.cn/problem-sets/994805342720868352/problems/994805472333250560 题意: 有k张乒乓球桌,有的是vip桌 ...

  5. PAT A1026 Table Tennis (30 分)——队列

    A table tennis club has N tables available to the public. The tables are numbered from 1 to N. For a ...

  6. PAT 1026 Table Tennis[比较难]

    1026 Table Tennis (30)(30 分) A table tennis club has N tables available to the public. The tables ar ...

  7. Pat(Advanced Level)Practice--1026(Table Tennis)

    Pat1026代码 题目描写叙述: A table tennis club has N tables available to the public. The tables are numbered ...

  8. PAT 1026. Table Tennis

    A table tennis club has N tables available to the public.  The tables are numbered from 1 to N.  For ...

  9. PAT 1026 Table Tennis (30)

    A table tennis club has N tables available to the public. The tables are numbered from 1 to N. For a ...

随机推荐

  1. 关于python3字符串中引号格式的看法,‘’,“”

    关于字符串中引号格式的看法 在python3中,字符串统一使用: ' ' 或  " "     来表示,两者没有什么区别. 但是在使用的过程中,可以注意一下使用的方式,可以提高编程 ...

  2. HDFS under replicated blocks

    under replicated blocks 解决: 找出没有复制的block: hdfs fsck / | grep 'Under replicated' | awk -F':' '{print ...

  3. 04_Mybatis输入\出映射

    1. 输入映射 ​ 通过paramterType指定输入参数的类型,类型可以是简单类型.hashmap.pojo的包装类. 1.1 传递pojo的包装对象 1.需求 ​ 完成用户信息的综合查询,需要传 ...

  4. 矩阵连乘 /// 区间DP oj1900

    题目大意: 输入t :t为测试用例个数 接下来t个测试 每个测试用例 第一行输入n: n为矩阵个数 保证n个矩阵依序是可乘的 接下来n行 每行输入p,q:p为长度q为宽度 对给定的n个矩阵确定一个计算 ...

  5. AM8 自定义表情包的实现方法

    AM8 自定义表情包的实现方法 效果描述 AM8 安装后,在\Activesoft\AMm8\emotions 目录内存储的是默认的表情符号.但有的时候我们需要增加一些新的表情符号,AM8 系统支持自 ...

  6. [转]Entity Framework 的实体关系

    通过 Entiy Framework实践系列 文章,理了理 Entity Framework 的实体关系. 为什么要写文章来理清这些关系?“血”的教训啊,刚开始使用 Entity Framework  ...

  7. Swimming Balls

    Swimming Balls https://vjudge.net/contest/318752#problem/J如果直接算,各种球的情况都不清楚,因为放一个球之后,水位的变化也会影响之前放入的球, ...

  8. SG函数模板(洛谷2197nim游戏

    #include <iostream> #include <cstdio> #include <queue> #include <algorithm> ...

  9. 2016.8.19上午初中部NOIP普及组比赛总结

    2016.8.19上午初中部NOIP普及组比赛总结 链接:https://jzoj.net/junior/#contest/home/1338 这次总结发得有点晚啊!我在这里解释一下, 因为浏览器的问 ...

  10. Tensorflow入门篇

     参考Tensorflow中文网(http://www.tensorfly.cn/tfdoc/get_started/introduction.html) ,写一个入门. 1.打开pyCharm,新建 ...