PAT甲级——A1026 Table Tennis
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 ;
}
最新文章
- PHP日志扩展 SeasLog-1.6.8, 性能更优
- nyoj 56-阶乘因式分解（一）
- sql server 2008 评估期已过期解决办法
- UEditor-JSP版部署说明
- juce中的内存泄漏检测
- php判断变量是否存在
- IOS开发-Swift新语言初见
- 无法删除MySql数据库，报错1010 error dropping
- 关于Java的静态：静态类、静态方法、静态变量、静态块等
- opencv 离线文档下载地址在哪里？
- bouncing-balls-evil-circle
- 使用shiro框架，解决跳转页面出现404的问题
- Python 3下Matplotlib画图中文显示乱码的解决方法
- maven 跳过test
- idea springboot热部署无效问题
- vc++使用cookie登录网站
- JwtBearerAppBuilderExtensions.UseJwtBearerAuthentication(IApplicationBuilder
- c++工厂模式(Factory method)
- 决策树算法（ID3）
- 虚拟机和宿主机不能互ping的解决办法等