C# Example analysis of using dynamic programming to solve 0-1 backpack problem

// Use dynamic programming to solve the 0-1 backpack problemusing System;
using ;
using ;
using ;
namespace Knapsack_problem
// The key to the backpack problem is to calculate the maximum value of the total capacity of the backpack.{
 class Program
 {
  static void Main()
  {
   int i;
   int capacity = 16;
   int[] size = new int[] { 3, 4, 7, 8, 9 };
   // Each of the 5 items is 3, 4, 7, 8, 9 in size respectively   //And it is inseparable 0-1 Backpacking problem   int[] values = new int[] { 4, 5, 10, 11, 13 };
   // Each of the 5 items is 4, 5, 10, 11, 13 respectively   int[] totval = new int[capacity + 1];
   // Array Totval is used to store the maximum total value   int[] best = new int[capacity + 1];
   // Best stores the most valuable items currently   int n = ;
   for (int j = 0; j <= n - 1; j++)
    for (i = 0; i <= capacity; i++)
     if (i >= size[j])
      if (totval[i] < (totval[i - size[j]] + values[j]))
   // If the current capacity minus J's capacity plus J's value is greater than the original value,   // Pass this value to the current value      {
       totval[i] = totval[i - size[j]] + values[j];
       best[i] = j; // and pass j to best      }
   ("The greatest value of a backpack: " + totval[capacity]);
   // ("The items that make up the backpack are: ");   // int totcap = 0;
   // while (totcap <= capacity)
   // {
   // ("The size of the item is:" + size[best[capacity - totcap]]);   //  for (i = 0; i <= n-1; i++)
   //  totcap += size[best[i]];
   // }
  }
 }
}