$classA = [ new Point(1,1),new Point(1,2),new Point(3,3),5) ];
$classB = [ new Point(4,new Point(5,new Point(4,new Point(6,6) ];

$pair1 = [$a[0],$b[0]];
$pair2 = [$a[1],$b[1]];

$classA = [ new Point(1,3)];
$classB = [ new Point(4,1)];

$combinations = [];
$pair1 = [];
$pair2 = [];

$n = count($classA);
$m = count($classB);

for ($i=0; $i<$n; $i++){

    for ($j=0; $j<$m; $j++){

        $pair1 = [ $classA[$i],$classB[$j] ];

        for ($z=0; $z<$n; $z++){

            if($z != $i && $z != $j){
                for ($y=0; $y<$m; $y++){
                    if($y != $i &amp;& $y != $j){

                        $pair2 = [ $classA[$z],$classB[$y] ];
                        $combinations[] = [$pair1,$pair2];

                    }
                }
            }

        }
    }
}

我建议首先创建所有可能的组合,然后对于每一对,在“所有可能的组合”中用null-s替换行/列,虽然它可能不是最快的,但它应该工作并且给你一般理念

$allCombinations = Array();
foreach($classA as $value) {
    $column = Array();
    foreach($classB as $bPart) {
        $column[] = Array($bPart,$value);
    }
    $allCombinations[] = $column;
}

$possibleCombinations = Array();

$sizeA = count($classA);
$sizeB = count($classB);

for($a = 0; $a < $sizeA; $a++) {
    $column = Array();
    for($b = 0; $b < $sizeB; $b++) {
        $temp = $allCombinations;

        for($i = 0;$i < $sizeA;$i++) {
            $temp[$a][$i] = null;
        }

        for($i = 0;$i < $sizeB;$i++) {
            $temp[$i][$b] = null;
        }       

        // for $classA[$a],$classB[$b] possible combinations are in temp Now
        $column[] = $temp;
    }
    $possibleCombinations[] = $column;
}

现在,在$possibleCombinations中,您可以看到给定A / B索引的可能组合

在第一个循环中,它只是创建所有可能的组合,
在第二个循环中,它将首先得到A和B值,复制所有组合数组,然后将A列和B行值设置为null(因为它们不能使用,对吧？),最后这个可能的组合存储在$temp中并添加到$possibleCombinations,

例：

$ab = Array($classA[$i],$classB[$j]);
$possibleCombinationsForAB = $possibleCombinations[$i,$j];