特定背包算法 [英] specific Knapsack Algorithm

问题描述

我在解决特定的背包算法问题时遇到问题。有没有人给我一些提示或帮助我？我用蛮力方法解决了这个问题，但是执行时间却很长（我检查了所有可能的组合并采取了最佳解决方案-可行）。我需要通过动态编程或贪婪算法（但通过DP更好）来解决它。我读了很多，但找不到解决方法； /这是艰苦的练习。
这里是我对锻炼的描述

I have problem with solve the specific knapsack algorithm problem. Is there someone who give me some tips or help me? I solved it by Brute Force method but the execute time is very long (I checked all possible combinations and take the best solution - it works). I need to solve it by Dynamic Programming or Greedy Algorithm (but better by DP). I read about it a lot and I can't find the solution with it ;/ It is hard exercise. HERE IS description of my exercise

此锻炼在这里测试

推荐答案

互联网上有一些很好的教程，可以彻底解释背包问题。

There are a few good tutorials on the internet that explain the Knapsack problem thoroughly.

更具体地说，我会推荐此特定的，其中完整解释了问题和DP方法，包括使用三种不同语言（包括Java）的解决方案。

More specifically, I would recommend this specific one, where the problem and the DP-approach is entirely explained, including the solution in three different languages (including Java).

// A Dynamic Programming based solution for 0-1 Knapsack problem
class Knapsack
{
    // A utility function that returns maximum of two integers
    static int max(int a, int b) { return (a > b)? a : b; }

   // Returns the maximum value that can be put in a knapsack of capacity W
    static int knapSack(int W, int wt[], int val[], int n)
    {
         int i, w;
     int K[][] = new int[n+1][W+1];

     // Build table K[][] in bottom up manner
     for (i = 0; i <= n; i++)
     {
         for (w = 0; w <= W; w++)
         {
             if (i==0 || w==0)
                  K[i][w] = 0;
             else if (wt[i-1] <= w)
                   K[i][w] = max(val[i-1] + K[i-1][w-wt[i-1]],  K[i-1][w]);
             else
                   K[i][w] = K[i-1][w];
         }
      }

      return K[n][W];
    }

    // Driver program to test above function
    public static void main(String args[])
    {
        int val[] = new int[]{60, 100, 120};
        int wt[] = new int[]{10, 20, 30};
        int  W = 50;
        int n = val.length;
        System.out.println(knapSack(W, wt, val, n));
    }
}
/*This code is contributed by Rajat Mishra */

来源： GeeksForGeeks

这篇关于特定背包算法的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！