PHP dynamic planning to solve 0-1 knapsack problem case Analysis, 0-1 example Analysis _php tutorial

PHP Dynamic planning solves 0-1 knapsack problem case analysis, 0-1 instance analysis

In this paper, the dynamic programming of PHP is analyzed to solve 0-1 knapsack problem. Share to everyone for your reference. The specific analysis is as follows:

Backpack problem Description: A backpack with a maximum weight of W, now has n items, each item weighs t, and the value of each item is v.
To make the backpack the most weight (but not more than W), but also need the most valuable backpack.

Idea: Define a two-dimensional array, one-dimensional for the number of items (for each item), two-dimensional is the weight (not exceeding the maximum, here is 15), the following array a,
The principle of dynamic programming, Max (OPT (i-1,w), wi+opt (I-1,W-WI)), the maximum value,
Opt (I-1,W-WI) refers to the previous optimal solution

<?php//This is what I wrote according to the principle of dynamic programming//MAX (opt (i-1,w), wi+opt (I-1,W-WI)//backpack can be installed maximum weight $w=15;//There are four items, weight per item $dx=array (3,4,5,6) ;//value of each item $qz=array (8,7,4,9);//define an array $a=array ();//Initialize for ($i =0; $i <=15; $i + +) {$a [0][$i]=0;} for ($j =0; $j <=4; $j + +) {$a [$j][0]=0;} Opt (i-1,w), wi+opt (I-1,W-WI) for ($j =1, $j <=4; $j + +) {for  ($i =1; $i <=15; $i + +) {    $a [$j] [$i]= $a [$j -1][$i ];    No greater than the maximum w=15    if ($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) = Compare       if ($tmp > $a [$j] [$i]) {        $a [$j] [$i]= $tmp      ;    }  }} To print this array, the right-most value of the output is the maximum value for ($j =0; $j <=4; $j + +) {for  ($i =0; $i <=15; $i + +) {    echo $a [$j] [$i]. " /t ";    } echo "/n";}? >

I hope this article is helpful to everyone's PHP programming.

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