Ask everyone about a deformed backpack.

Ask everyone about a deformed backpack. there are N kinds of items and a backpack with a capacity of V. There is also a threshold T.
A maximum of n [I] items are available for the I-th item. the cost per item is c [I] and the value is w [I].
Solving which items are loaded into a backpack can make the total cost of these items not exceed the capacity of the backpack, and the total value is greater than and closest to T.

Complete backpack code

/**
* Knapsack problem description: a backpack with a maximum weight of W now has n items, each weighing t, and the value of each item is v.
* To maximize the weight of a backpack (but not more than W) and maximize the value of a backpack.
* Idea: define a two-dimensional array. the one-dimensional array is the number of items (indicating each item), and the two-dimensional array is the weight (no more than the maximum, here is 10). The following array is,
* The principle of dynamic planning, max (opt (i-1, w), wi + opt (i-1, w-wi) among the maximum,
* Opt (i-1, w-wi) refers to the previous optimal solution
*/
// This is based on the dynamic planning principle.
// Max (opt (i-1, w), wi + opt (i-1, w-wi ))
// The maximum weight of a backpack
$ W = 10;
// There are four items, the weight of each item
$ Dx = array (3, 4, 5 );
// Value of each item
$ Qz = array (4, 5, 6 );
// Define an array
$ A = array ();
// Initialization
For ($ I = 0; $ I <= 10; $ I ++) {$ a [0] [$ I] = 0 ;}
For ($ j = 0; $ j <= 3; $ j ++) {$ a [$ j] [0] = 0 ;}
// Opt (i-1, w), wi + opt (i-1, w-wi)
For ($ j = 1; $ j <= 3; $ j ++ ){
For ($ I = 1; $ I <= 10; $ I ++ ){
$ A [$ j] [$ I] = $ a [$ j-1] [$ I];
// W = 10
If ($ dx [$ j-1] <= $ w ){
If (! Isset ($ a [$ j-1] [$ I-$ dx [$ j-1]) continue;
// Wi + opt (i-1, wi)
$ Tmp = $ a [$ j-1] [$ I-$ dx [$ j-1] + $ qz [$ j-1];
// Opt (i-1, w), wi + opt (i-1, w-wi) => comparison
If ($ tmp> $ a [$ j] [$ I]) {
$ A [$ j] [$ I] = $ tmp;
}
}
}
}
// Print This array and output the rightmost value.
For ($ j = 0; $ j <= 3; $ j ++ ){
For ($ I = 0; $ I <= 10; $ I ++ ){
Echo $ a [$ j] [$ I]. "";
} Echo"
";
}


?>

Reply to discussion (solution)

What problems have you encountered? Your results should be correct.
But I like to write it like this.

$w = 10;$ar = array(  array('w' => 3, 'v' => 4),  array('w' => 4, 'v' => 5),  array('w' => 5, 'v' => 6),);foreach($ar as $k=>$v) {  $v['k'][] = $k;  $res[] = $v;}$p = 0;for(;$p
 
  $v) {    if(in_array($i, $res[$p]['k'])) continue;    if($r['w'] + $v['w'] <= $w) {      $res[] = array(        'w' => $r['w'] + $v['w'],        'v' => $r['v'] + $v['v'],        'k' => array_merge($r['k'], array($i)),      );    }  }}foreach($res as $v) $t[] = $v['v'];array_multisort($t, SORT_DESC, $res); print_r($res);
 

Array(    [0] => Array        (            [w] => 9            [v] => 11            [k] => Array                (                    [0] => 1                    [1] => 2                )        )    [1] => Array        (            [w] => 9            [v] => 11            [k] => Array                (                    [0] => 2                    [1] => 1                )        )    [2] => Array        (            [w] => 8            [v] => 10            [k] => Array                (                    [0] => 0                    [1] => 2                )        )    [3] => Array        (            [w] => 8            [v] => 10            [k] => Array                (                    [0] => 2                    [1] => 0                )        )    [4] => Array        (            [w] => 7            [v] => 9            [k] => Array                (                    [0] => 0                    [1] => 1                )        )    [5] => Array        (            [w] => 7            [v] => 9            [k] => Array                (                    [0] => 1                    [1] => 0                )        )    [6] => Array        (            [w] => 5            [v] => 6            [k] => Array                (                    [0] => 2                )        )    [7] => Array        (            [w] => 4            [v] => 5            [k] => Array                (                    [0] => 1                )        )    [8] => Array        (            [w] => 3            [v] => 4            [k] => Array                (                    [0] => 0                )        ))

At present, this is the greatest value of a backpack filled with w = 10.

I am trying to find out: a combination of a total value close to a given threshold T when it is as full as possible

$f = 9;print_r(array_filter($res, function($a) use ($f) { return $a['v'] > 0.9*$f && $a['v']<1.1*$f;}));

Thank you! Very good B, the result is completely achieved

This algorithm is too concise to understand. how can I filter out repeated combinations? For example, 0 3 6, 3 0 6, 6 0 3

.... Array_multisort ($ t, SORT_DESC, $ res); // sort (existing) foreach ($ res as $ I => $ v) if (isset ($ res [$ i-1]) &! Array_diff ($ res [$ i-1] ['K'], $ v ['K']) unset ($ res [$ I]); // deduplicate $ res = array_values ($ res); // normalization

I did not do this algorithm according to the dynamic plan.
Instead, all possible combinations are recorded and checked by sorting.

Thank you very much for your advice.