1026. Table Tennis (30)

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:

920:52:00 10 008:00:00 20 008:02:00 30 020:51:00 10 008:10:00 5 008:12:00 10 120:50:00 10 008:01:30 15 120:53:00 10 13 12

Sample Output:

08:00:00 08:00:00 008:01:30 08:01:30 008:02:00 08:02:00 008:12:00 08:16:30 508:10:00 08:20:00 1020:50:00 20:50:00 020:51:00 20:51:00 020:52:00 20:52:00 03 3 2

//做这个题目快做疯了，真不知道自己第一次是怎么想出来的，参考以前的代码重新写了一遍

#include <iostream>#include <cstdio>#include <vector>#include <algorithm>#include <set>#include <cmath>#include <climits>using namespace std;struct Play{  int start, p;  bool operator < (const Play &player) const  {    return start < player.start;  }};struct Table{  int freetime, number;  Table():freetime(0), number(0) {}};int MAXTIME = 13 * 3600;vector<Play> players;vector<Play> vplayers;set<int> vtable;vector<Table> table;void printTime(int time, int serve){  printf("%02d:%02d:%02d %02d:%02d:%02d %d\n", time / 3600 + 8, (time / 60) % 60, time % 60, serve / 3600 + 8, (serve / 60) % 60, serve % 60, (serve - time + 30) / 60);}int main(void){  int N, K, M, hour, minite, second, p, tag, vip;  scanf("%d", &N);  Play player;  for (int i = 0; i < N; ++i)  {    scanf("%d:%d:%d%d%d", &hour,&minite, &second,&p, &tag);    player.start = (hour - 8) * 3600 + minite * 60 + second;            player.p = p > 120 ? 7200 : p * 60;     if (tag == 1)      vplayers.push_back(player);    else      players.push_back(player);  }  scanf("%d%d", &K, &M);  table.resize(K);  for (int i = 0; i < M; ++i)  {    scanf("%d", &vip);    vtable.insert(--vip);  }  sort (players.begin(), players.end());  sort (vplayers.begin(), vplayers.end());  int length = players.size(), vlength = vplayers.size(), index1 = 0, index2 = 0;  int startime, servetime;  while(index1 < length || index2 < vlength)  {  //贵宾    if (index1 < length && index2 < vlength)    {      //如果有普通客户和贵宾，则找出时间来的最早那个      if (players[index1].start < vplayers[index2].start)      {        startime = players[index1].start;        tag = 0;      }      else      {        startime = vplayers[index2].start;        tag = 1;      }    }    if (index1 < length && index2 == vlength)    {      //只有普通客户      startime = players[index1].start;      tag = 0;    }    if (index1 == length && index2 < vlength)    {      //只有贵宾      startime = vplayers[index2].start;      tag = 1;    }if (startime >= MAXTIME) //用户来的时间都超过了营业时间break;    //1、寻找空着数字最小的席位； 2、如果没有空着的席位，寻找最早结束的席位    int minIndex = -1, minStart = INT_MAX;    if (tag)    {        for (set<int>::iterator iter = vtable.begin(); iter != vtable.end(); ++iter)      {        if (table[*iter].freetime <= startime)        {          minIndex = *iter;          break;        }      }    }    if (minIndex == -1)    {        for (int j = 0; j < K; ++j)      {        if (table[j].freetime <= startime)        {          minIndex = j;          break;        }        if (table[j].freetime < minStart)        {          minStart = table[j].freetime;          minIndex = j;        }      }      }    if (table[minIndex].freetime >= MAXTIME)      break;
<span style="white-space:pre"></span>//如果minIndex的时间比starttime小，并且它是贵宾桌，很可能它会分配给vip用户，这个用户必须满足这个贵宾桌freetime之前来到，否则它会直接非配给starttime。    if (index2 < vlength && vtable.find(minIndex) != vtable.end() && vplayers[index2].start <= table[minIndex].freetime)      tag = 1;        if (table[minIndex].freetime < startime)      servetime = startime;    else      servetime = table[minIndex].freetime;    ++table[minIndex].number;    if (tag)    {      table[minIndex].freetime = servetime + vplayers[index2].p;      printTime(vplayers[index2].start, servetime);      ++index2;    }    else    {      table[minIndex].freetime = servetime + players[index1].p;      printTime(players[index1].start, servetime);      ++index1;    }      }  printf("%d", table[0].number);  for (int i = 1; i < K; ++i)    printf(" %d", table[i].number);  printf("\n");  return 0;}