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
和之前做的银行排队服务的问题类似,区别在于这里增加了vip选项。如果有vip的台子空出来,在排队的人中有vip,那么就会将vip台子分给他;否则就分给普通人。
当前排队队首是vip,空出的台子不是vip台子,不会立即分配,而是寻找在时间范围内是否有vip台子可以空出来。
最初想分情况讨论,写的很复杂,时间还会超时。对于这种问题,可以这样来解决的:
依次找最先空出来的台子,在刚进来的人中,根据是否是vip,空出的台子是否为vip进行分配。
代码如下:
#include<iostream>
#include<cstdio>
#include<vector>
#include<algorithm>
#include<cmath>
using namespace std;
struct pla{
int arrive,start,p,tag;
};
struct tab
{
int end=*;
int vip,num;
};
vector<pla>player;
vector<tab>table;
const int inf=0x3f3f3f3f;
bool cmp1(pla a,pla b)
{
return a.arrive<b.arrive;
}
bool cmp2(pla a,pla b)
{
return a.start<b.start;
}
int findvip(int id)
{
id++;
while(id<player.size()&&player[id].tag!=)
id++;
return id;
}
void allocate(int person,int index)
{ if(player[person].arrive<=table[index].end)
player[person].start=table[index].end;
else
player[person].start=player[person].arrive;
//cout<<person<<" "<<index<<player[person].start<<" "<<player[person].arrive<<endl;
table[index].end=player[person].start+player[person].p;
table[index].num++;
}
int main()
{
int n;
scanf("%d",&n);
for(int i=;i<=n;i++)
{
int h,m,s,p,tag;
scanf("%d:%d:%d %d %d",&h,&m,&s,&p,&tag);
pla temp;
temp.arrive=h*+m*+s;
temp.start=*;
if(temp.arrive>=*) //是否不符合条件
continue;
temp.p=p<=?p*:; //测试用例有超过2小时的情况
temp.tag=tag;
player.push_back(temp);
}
int k,m;
scanf("%d%d",&k,&m);
table.resize(k+);
for(int i=;i<=m;i++)
{
int tmp;
scanf("%d",&tmp);
table[tmp].vip=;
}
sort(player.begin(),player.end(),cmp1);
int i=,vipid=-;
vipid=findvip(vipid); //寻找vip成员
while(i<player.size())
{
int index=-,minendtime=inf;
for(int j=;j<=k;j++) //寻找最早结束的桌子
{
if(table[j].end<minendtime)
{
minendtime=table[j].end;
index=j;
}
}
if(table[index].end>=*)
break;
if(player[i].tag==true&&i<vipid) //说明此vip已经分配过了
{
i++;
continue;
}
if(table[index].vip==)
{
if(player[i].tag==)
{
allocate(i,index);
if(vipid==i)
vipid=findvip(vipid);
i++;
}
else
{
if(vipid<player.size()&&player[vipid].arrive<=table[index].end)
{
allocate(vipid,index);
vipid=findvip(vipid);
}
else
{
allocate(i,index);
i++;
}
}
}
else
{
if(player[i].tag!=)
{
allocate(i,index);
i++;
}
else
{
int vipindex=-,vipendtime=inf;
for(int j=;j<=k;j++)
{
if(table[j].vip==&&table[j].end<vipendtime)
{
vipindex=j;
vipendtime=table[j].end;
}
}
if(vipindex!=-&&player[i].arrive>=table[vipindex].end)
{
allocate(i,vipindex);
if(i==vipid)
vipid=findvip(vipid);
i++;
}
else
{
allocate(i,index);
if(i==vipid)
vipid=findvip(vipid);
i++;
} }
}
}
sort(player.begin(),player.end(),cmp2);
for(int 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<table.size();i++)
{
if(i!=)
printf(" ");
printf("%d",table[i].num);
}
}

 

PAT1026 (大模拟)的更多相关文章

  1. HDU 5920 Ugly Problem 高精度减法大模拟 ---2016CCPC长春区域现场赛

    题目链接 题意:给定一个很大的数,把他们分为数个回文数的和,分的个数不超过50个,输出个数并输出每个数,special judge. 题解:现场赛的时候很快想出来了思路,把这个数从中间分为两部分,当位 ...

  2. AC日记——神奇的幻方 洛谷 P2615(大模拟)

    题目描述 幻方是一种很神奇的N*N矩阵:它由数字1,2,3,……,N*N构成,且每行.每列及两条对角线上的数字之和都相同. 当N为奇数时,我们可以通过以下方法构建一个幻方: 首先将1写在第一行的中间. ...

  3. ACdream 1188 Read Phone Number (字符串大模拟)

    Read Phone Number Time Limit:1000MS     Memory Limit:64000KB     64bit IO Format:%lld & %llu Sub ...

  4. 2016ACM-ICPC网络赛北京赛区 1001 (trie树牌大模拟)

    [题目传送门] 1383 : The Book List 时间限制:1000ms 单点时限:1000ms 内存限制:256MB 描述 The history of Peking University ...

  5. Bzoj1972: [Sdoi2010]猪国杀 题解(大模拟+耐心+细心)

    猪国杀 - 可读版本 https://mubu.com/doc/2707815814591da4 题目可真长,读题都要一个小时. 这道题很多人都说不可做,耗时间,代码量大,于是,本着不做死就不会死的精 ...

  6. (大模拟紫题) Luogu P1953 易语言

    原题链接:P1953 易语言 (我最近怎么总在做大模拟大搜索题) 分别处理两种情况. 如果只有一个1或0 直接设一个cnt为这个值,每次输入一个新名字之后把数字替换成cnt,最后cnt++即可. 注意 ...

  7. NOIP2017 时间复杂度 大模拟

    再写一道大模拟题. 由于是限时写的,相当于考场代码,乱的一批. 题目链接:P3952 时间复杂度 先记几个教训: 字符串形式的数字比较大小老老实实写函数,字典序都搞错几次了 栈空的时候不但pop()会 ...

  8. [CSP-S模拟测试]:引子(大模拟)

    题目描述 网上冲浪时,$Slavko$被冲到了水箱里,水箱由上而下竖直平面.示意图如下: 数字$i$所在的矩形代表一个编号为$i$的水箱.1号水箱为水箱中枢,有水管连出.除了$1$号水箱外,其他水箱上 ...

  9. 模拟赛38 B. T形覆盖 大模拟

    题目描述 如果玩过俄罗斯方块,应该见过如下图形: 我们称它为一个 \(T\) 形四格拼板 .其中心被标记为\(×\). 小苗画了一个 \(m\) 行 \(n\) 列的长方形网格.行从 \(0\) 至 ...

随机推荐

  1. golang判断文件/文件夹是否存在

    使用os包,os.stat返回err==nil,说明存在: os.IsNotExist(err)为true,说明不存在:否则不确定是否存在 func DelJar(fileName string) e ...

  2. 推荐一款idea 翻译插件 ECTranslation

    无意中看到一款idea翻译插件, ECTranslation,才知道有这么个东西,推荐给看到的人吧,使用简单,值得拥有. 参考:http://p.codekk.com/detail/Android/S ...

  3. js -history.back(-1)和history.go(-1) 区别

    既然history.back(-1)和history.go(-1)都是返回之前页面,   history.back(-1)//直接返回当前页的上一页,,是个新页面   history.go(-1)// ...

  4. supervisor process management

    supervisor是unix like系统的进程管理工具 安装: pip install supervisor 配置文件: echo_supervisord_conf # 打印一个配置文件样例 ec ...

  5. html:块级元素和行内元素的特点

    display:block: 块元素会独自占据一整行,或者多行,可以任意设置其大小尺寸,是用于搭建网页布局的必须部分,使网页结构更加紧凑合理. 块级元素width.height.padding.mar ...

  6. C++复习:纯虚函数和抽象类

    纯虚函数和抽象类 1基本概念 2抽象类案例   3抽象类在多继承中的应用 C++中没有Java中的接口概念,抽象类可以模拟Java中的接口类.(接口和协议) 3.1有关多继承的说明 工程上的多继承 被 ...

  7. jsp开发环境搭建(windows64位)

    有些东西当时学和用的时候很熟练,但如果时间久了不用了,再次遇到的时候,也会很生疏,现在对一般的jsp网站开发环境的搭建做一个小结,以备以后不时之需,作为参考手册用. 一.java环境搭建 1.下载jd ...

  8. ReactiveX 学习笔记(12)调度器

    Schedulers, threading and testing 本文的主题为调度器,多线程以及测试. RxJava操作符(六)Utility SubscribeOn / ObserveOn Sub ...

  9. django 认证系统--2

    使用django的认证系统 User 对象 User是认证系统的核心.典型代表是用户和你的站点进行交互还有限制访问.注册用户等等.django认证框架中,只存在一个User类,像'superuser' ...

  10. Structs复习 Action

    引入jar包 web.xml <?xml version="1.0" encoding="UTF-8"?> <web-app version= ...