java – 选择0/1背包中的物品,其中两个物品具有相同的好处|最大

发布时间：2020-12-15 02:08:01 所属栏目：Java 来源：网络整理

导读：在0/1背包问题中,如果两个项目具有相同的值,如何选择项目.应选择体重较轻的值,如何检查该状况？我使用动态编程有以下功能. static int[] knapsack(int maxWeight,double[] weight,double[] value,int n) { //n = no. if items int i,w; double array[][] = n

在0/1背包问题中,如果两个项目具有相同的值,如何选择项目.应选择体重较轻的值,如何检查该状况？我使用动态编程有以下功能.

static int[] knapsack(int maxWeight,double[] weight,double[] value,int n) {
    //n = no. if items
    int i,w;

    double array[][] = new double[n + 1][maxWeight + 1];
    for (i = 0; i <= n; i++) {
        for (w = 0; w <= maxWeight; w++) {
            if (i == 0 || w == 0)
                array[i][w] = 0;
            else if (weight[i - 1] <= w)
                array[i][w] = max(value[i - 1] + array[i - 1][(w -(int) weight[i - 1])],array[i - 1][w]);
            else
                array[i][w] = array[i - 1][w];
            if (i != 0 || w != 0)
                System.out.print(array[i][w] + "t");
        }
        System.out.println();
    }

    int[] selected = new int[n + 1];
    for (int j = n,wt = maxWeight; j > 0; j--) {   
        if (array[j][wt] != array[j - 1][wt]) {
            if (array[j][wt] == array[j][wt - 1]) {
                selected[j] = 0;
                break;
            }
            selected[j] = 1;
            wt = wt - (int) weight[j - 1];
        }
        else
            selected[j] = 0;
    }
    /** Print finally selected items **/
    System.out.println("nItems selected : ");
    for (int k = 1; k < n + 1; k++)
        if (selected[k] == 1)
            System.out.print(k +" ");
        System.out.println();

        return selected;
}

对于这种情况：(i,v):( 4,45)(3,20)(5,30)(2,45),maxWeight = 5;
?如果第1项和第4项具有相同的值,则应选择权重较小的第4项.如何在上面的代码中实现这个条件.
问题陈述：

Your goal is to determine which things to put into the
package so that the total weight is less than or equal to the package
limit and the total cost is as large as possible. You would prefer to
send a package which weights less in case there is more than one
package with the same price.

解决方法

如果您的意思是最大限度地减少重量值.你可以检查一下

设DP [i] [j]是可以获得的最大值！

并且W [i] [j]要使用的最小重量!!

然后,

if(Current Weight > Element's Weight)
{
      if(DP[i-1][j-Weight[i]]+Value[i]>DP[i-1][j]){
             DP[i][j]=DP[i-1][j-Weight[i]]+Value[i];
             Weight[i][j]= Weight[i-1][j-Weight[i]]+Value[i]
      }
      else if(DP[i-1][j-Weight[i]]+Value[i] <  DP[i-1][j] ){
             DP[i][j]=DP[i-1][j];
             Weight[i][j]=Weight[i-1][j];
      } 
      else{                   //Note this is the tricky part elsewise the 
                              //Above conditions are simple Knapsack conditions

             DP[i][j]=DP[i-1][j];   //Both of them are equal We will get same Value . Thus we cannot maximise it in any other way!!
             Weight[i][j]=minimum ( Weight[i-1][j],Weight[i-1][j-Weight[i]]+A[i]);

      }
}
else
{
             DP[i][j]=DP[i-1][j];
             Weight[i][j]=Weight[i-1][j];
}

注意解决方案是微不足道的,除非第一个条件中的第三个条件！
我们需要不惜一切代价最大化乐趣！所以我们不要惹它！
但是当两种情况下的乐趣相同时,我们需要选择重量较轻的那种我们最终会为相同的背包价值增加重量！

我假设你知道背包0/1的问题,这就是为什么我没有解释第一和第二个条件！

（编辑：李大同）

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