PHP dynamic planning solution for 0-1 knapsack problem instance analysis

PHP dynamic planning solution for 0-1 knapsack problem instance analysis

This article mainly introduces PHP dynamic planning to solve the 0-1 knapsack problem. The example analyzes the principle and implementation skills of the knapsack problem. For more information, see

 

 

This article analyzes the PHP dynamic planning to solve the 0-1 knapsack problem. Share it with you 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 with a weight of t and the value of each item is v.
To maximize the weight of this backpack (but not more than W), and at the same time, the greatest value of the backpack is required.

Idea: define a two-dimensional array, one-dimensional is the number of items (indicating each item), two-dimensional is the weight (no more than the maximum, here is 15), the following array,
Max (I-1, w), wi + opt (I-1, w-wi,
Opt (I-1, w-wi) refers to the previous Optimal Solution

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<? Php // 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 = 15; // There are four items, the weight of each item $ Dx = array (3, 4, 5, 6 ); // Value of each item $ Qz = array (8, 7, 4, 9 ); // Define an array $ A = array (); // Initialization 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]; // 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) => comparison If ($ tmp> $ a [$ j] [$ I]) { $ A [$ j] [$ I] = $ tmp; } } } } // Print this array and output the rightmost value. For ($ j = 0; $ j <= 4; $ j ++ ){ For ($ I = 0; $ I <= 15; $ I ++ ){ Echo $ a [$ j] [$ I]. "/t "; } Echo "/n "; } ?>

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