如何配对数组中的项目？ 让我们说我有一系列战士。我想根据他们的权重来配对它们。最接近重量的战士应配对为最佳匹配。但如果他们在同一个团队中他们就不应该配对。
输出：
我一直在研究这个主题，发现类似但不完全的东西： Random But Unique Pairings, with Conditions
非常感谢一些帮助。提前谢谢！
答案 0 :(得分：5)
我很喜欢你的问题，所以我制作了一个完整的版本。
<?php
header("Content-type: text/plain");
error_reporting(E_ALL);
/**
* @class Fighter
* @property $name string
* @property $weight int
* @property $team string
* @property $paired Fighter Will hold the pointer to the matched Fighter
*/
class Fighter {
public $name;
public $weight;
public $team;
public $paired = null;
public function __construct($name, $weight, $team) {
$this->name = $name;
$this->weight = $weight;
$this->team = $team;
}
}
/**
* @function sortFighters()
*
* @param $a Fighter
* @param $b Fighter
*
* @return int
*/
function sortFighters(Fighter $a, Fighter $b) {
return $a->weight - $b->weight;
}
$fighterList = array(
new Fighter("A", 60, "A"),
new Fighter("B", 65, "A"),
new Fighter("C", 62, "B"),
new Fighter("D", 60, "B"),
new Fighter("E", 64, "C"),
new Fighter("F", 66, "C")
);
usort($fighterList, "sortFighters");
foreach ($fighterList as $fighterOne) {
if ($fighterOne->paired != null) {
continue;
}
echo "Fighter $fighterOne->name vs ";
foreach ($fighterList as $fighterTwo) {
if ($fighterOne->team != $fighterTwo->team && $fighterTwo->paired == null) {
echo $fighterTwo->name . PHP_EOL;
$fighterOne->paired = $fighterTwo;
$fighterTwo->paired = $fighterOne;
break;
}
}
}
usort()和排序函数sortFighters()按每个元素的weight属性排序。$fighterVariable->paired访问每个战士的对）答案 1 :(得分：2)
这只是基于评论的Truth's答案的延伸：
我做的第一件事就是基本保持跟踪球员。
$unassignedPlayers = $fighterList;
算法会起作用：准备团队列表（如果您正在使用数据库，请使用SELECT DISTINCT或GROUP BY teams.id）：
$teams = array();
foreach( $fighterList as $fighter){
$teams[] = $figter->team;
}
$teams = array_unique( $teams);
接下来我们需要一种能够分解战士阵营的方法（比方说，我们有团队{A,A,B,B,C,C}，我们希望将其分成{A,A}，{B,B,C,C}）：
// Don't use string type declaration, it's just ilustrating
function splitFighters( array $input, string $team){
$inteam = array();
$outteam = array();
foreach( $input as $fighter){
if( $figter->team == $team){
$inteam[] = $fighter;
} else {
$outteam[] = $fighter;
}
}
return array( $inteam, $outteam);
}
现在我们确实拥有了这个功能，我们可以创建一个对团队成员进行排序的功能：
function assignFighters( array &$input, array $teams, array &$output){
// Nothing to work with?
if( !count( $input)){
return true;
}
// No team left and still unassigned players, that fatal error
if( !cont( $teams)){
throw new Exception( 'Unassigned players occurred!');
}
// Shift team
$team = array_shift( $teams);
// Split into in and out team
list( $inteam, $outteam) = splitFighters( $input, $team);
// Inteam is already empty (let's say all players were assigned before)
// Just go deeper (where's DiCaprio?)
if( !count( $inteam) && count( $teams)) {
return assignFighters( $input, $teams, $output)
}
// There're unassigned and nonassignable players in this team
// This is error and we'll have to deal with it later
if( !count($outteam)){
$input = $inteam; // Propagate unassigned players to main
return false;
}
// Sort both teams by fighters weight
// Uses Truth's comparison function
usort($inteam, "sortFighters");
usort($outteam, "sortFighters");
// Fig = Fighter
while( $fig1 = array_shift( $inteam)){
// Are there any players to work with
if( !count( $outteam)){
array_unshift( $inteam, $fig1);
$input = $inteam; // Propagate unassigned players to main
return false;
}
// Assign players to each other
$fig2 = array_shift( $outteam);
$fig1->paired = $fig2;
$fig2->paired = $fig1;
// This will propagate players to main nicely
$output[] = $fig1;
$output[] = $fig2;
}
// Still here? Great! $inteam is empty now
// $outteam contains all remaining players
$input = $outteam;
return assignFighters( $input, $teams,$output);
}
在此之前你可以使用Truth的算法，但这应该有更好的权重匹配，并且更清楚地表示你想要的内容，但无论如何现在$unassignedPlayers进入工作：
$assignedPlayers = array();
$state = assignFighters( $unassignedPlayers, $teams, $assignedPlayers);
// Note:
$state === !(bool)count($unassignedPlayers)
// should evaluate as true, otherwise I'm having an error in algorithm
那么现在......如果你有$state === false resp。 count( $unassignedPlayers) > 0出了点问题我们需要应用一些魔法。这魔术将如何发挥作用：
// Keep the trace of swapped players so we don't end up in endless loop
$swappedPlayers = array();
// Browse all unassigned players
while( $fig1 = array_shift( $unassignedPlayers)){
// Store as swapped
$swappedPlayers[] = $fig1;
// At first check whether there's not unassigned player in the different team
// this shouldn't occur in first iteration (all fighters should be from one team
// in the beginning) but this is most effective part of this method
foreach( $unassignedPlayers as $key => $fig2){
if( $fig2->team != $fig1->team){
$fig1->pair = $fig2;
$fig2->pair = $fig1;
continue;
}
}
// No luck, normal magic required
list( $inteam, $outteam) = splitFighters( $assignedPlayers, $fig1->team);
$fig2 = null; // I like my variables initialized, this actually quite important
// Now select someone from $outteam you will want to swap fights with.
// You may either iterate trough all players until you find best weight
// match or select it random, or whatever, I'll go with random,
// it's your job to implement better selection
$i = 1000; // Limit iterations
while(($i--) > 1){
$key1 = array_rand( $outteam, 1);
if( $outteam[$key]->team == $fig1->team){
continue; // No point in swapping fit team member
}
// No recursive swaps
if( in_array( $outteam[$key], $swappedPlayers)){
continue;
}
// This may speed things really up:
// That means we'll get rid of 2 players due to foreach loop at the beggining
// However I'm not sure how this condition will really work
if( $outteam[$key]->pair->team == $fig1->team){
continue;
}
// Store matched fighter
$fig2 = $outteam[$key];
// Unset pair from another fighter
$fig2->pair->pair = null;
// Find the pair in $assignedPlayers and move it to $unassignedPlayers
$key = array_search( $fig2->pair, $assignedPlayers);
if( $key === false){
throw new Exception( 'Cannot find pair player');
}
unset( $assignedPlayers[$key]);
$unassignedPlayers[] = $fig2->pair;
$swappedPlayers[] = $fig2->pair;
// Remove pair from self
$fig2->pair = null;
$swappedPlayers[] = $fig2;
break; // hh, try forgetting this one :)
}
// This shouldn't be happening
if( $fig2 === null){
throw new Exception( 'Didn\'t find good match in 1000 iterations.');
}
// Ok now just make matches as go to the next iteration
$fig1->pair = $fig2;
$fig2->pair = $fig1;
// And store those
$assignedPlayers[] = $fig1;
$assignedPlayers[] = $fig2;
}
我已经把所有这些都写下来了（这是挑战），测试它并在评论中留下注释：）
答案 2 :(得分：0)
按重量对数组进行排序。然后，您将拥有彼此接近的权重对。