需要邏輯幫助，製作三維數組（PHP）

Question

我需要幫助，例如，我有4個隊列名的數組。

 
array('logiX.L4d22','Lust','Marat and Friends','Pandas of Belgium'); 

我想使3維數組，其中第一個是圓的，第二是比賽，第三個是誰的球隊與海誓山盟玩，始終存在一些2隊。

邏輯必須讓所有的球隊必須與所有其他球員一起打球，並且在一輪中，任何球隊只會打一場比賽，所以如果我們有5支球隊，那麼在某輪比賽中，一支球隊必須等待下一輪。

它必須產生這樣的：

 
array 
    0 => 
    array 
     0 => 
     array 
      0 => string 'logiX.L4D2' (length=10) 
      1 => string 'Lust' (length=4) 
     1 => 
     array 
      0 => string 'Marat and Friends' (length=17) 
      1 => string 'Pandas of Belgium' (length=17) 
    1 => 
    array 
     0 => 
     array 
      0 => string 'logiX.L4D2' (length=10) 
      1 => string 'Marat and Friends' (length=17) 
     1 => 
     array 
      0 => string 'Lust' (length=4) 
      1 => string 'Pandas of Belgium' (length=17) 
    2 => 
    array 
     0 => 
     array 
      0 => string 'logiX.L4D2' (length=10) 
      1 => string 'Pandas of Belgium' (length=17) 
     1 => 
     array 
      0 => string 'Lust' (length=4) 
      1 => string 'Marat and Friends' (length=17) 

它必須是工作，2,3,5 ...... 10 ...... 12支球隊。

我希望你能幫助我，我已經花了1天半的時間。

Answer 1

上循環算法PHP一些谷歌搜索提供了以下：

http://speedtech.it/blog/2009/03/15/round-robin-algorithm-php/

http://www.phpbuilder.com/board/showthread.php?t=10300945

我希望你會發現你在找什麼。

編輯

添加我爲這樣的嘗試，繼round-robin algorithm described on Wikipedia。

如果團隊編號爲奇數，則會在數組中添加一個團隊（空值），因此您可以檢索每輪的「等待團隊」。

<?php 

$teams = range('a', 'g'); 

function make_rounds($teams) 
{ 
    $nb_teams = count($teams); 

    if ($nb_teams % 2 != 0) 
    { 
    $teams[] = null; 
    $nb_teams++; 
    } 

    $nb_rounds = $nb_teams - 1; 
    $nb_matches = $nb_teams/2; 

    $rounds = array(); 

    for($round_index = 0; $round_index < $nb_rounds; $round_index++) 
    { 
    $matches = array(); 

    for($match_index = 0; $match_index < $nb_matches; $match_index++) 
    { 
     if ($match_index == 0) 
     $first_team = $teams[0]; 
     else 
     $first_team = $teams[(($nb_teams-2) + $match_index - $round_index) % ($nb_teams-1) + 1]; 

     $second_team = $teams[(($nb_teams*2) - $match_index - $round_index - 3) % ($nb_teams-1) + 1]; 

     $matches[] = array($first_team, $second_team); 
    } 

    $rounds[] = $matches; 
    } 

    return $rounds; 
} 

print_r(make_rounds($teams)); 

Answer 2

我的解決方案版本。我會稱之爲蠻力。但是，它以某種方式起作用。 （或者看起來就是這樣。）

<?php 

$a = array('a','b','c','d','e','f','g'); 

function do_array($a) 
{ 
    $lim = sizeof($a) - 1; 

    # Create an array of all matches to play. 
    $cross = array(array()); 
    foreach (range(0,$lim) as $k_row): 
     foreach (range(0,$lim) as $k_col): 
      if ($k_row >= $k_col): 
       $toput = false; 
      else: 
       $toput = array($a[$k_row],$a[$k_col]); 
      endif; 
      $cross[$k_row][$k_col] = $toput; 
     endforeach; 
    endforeach; 

    $ret = array(); 
    foreach (range(0,$lim) as $k_round): 
     $round = array(); 
     # $tmp array holds all possible matches 
     # to play in current round. 
     $tmp = $cross; 
     $i = 0; 
     foreach (range(0,$lim) as $k_row): 
      foreach (range(0,$lim) as $k_col): 
       if ($math = $tmp[$k_row][$k_col]): 
        $cross[$k_row][$k_col] = false; 
        # These matches are not possible 
        # in the current round. 
        foreach (range(0,$lim) as $k): 
         $tmp[$k][$k_col] = false; 
         $tmp[$k][$k_row] = false; 
         $tmp[$k_col][$k] = false; 
         $tmp[$k_row][$k] = false; 
        endforeach; 
        $round[] = $math; 
       endif; 
      endforeach; 
     endforeach; 
     if ($round): 
      $ret[] = $round; 
     endif; 
    endforeach; 

    return $ret; 
} 

print_r (do_array($a)); 

?>