如何在0-1背包中获取所选物品的清单？ [英] How to get the list of selected items in 0-1 knapsack?
问题描述
我有一个背包问题天真解决方案的代码，我想获取所选项目的索引列表，当前它返回所选项目的值的总和。
任何帮助将不胜感激。
JAVA代码：
i have a code of the naive solution of the Knapsack problem, i want to get the list of index of selected items, currently it is returning the total sum of values of the selected items. Any help will be appreciated. JAVA CODE:
/* package whatever; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
/* A Naive recursive implementation of 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(float W, float wt[], int val[], int n)
{
// Base Case
if (n == 0 || W == 0)
return 0;
// If weight of the nth item is more than Knapsack capacity W, then
// this item cannot be included in the optimal solution
if (wt[n-1] > W)
{
return knapSack(W, wt, val, n-1);
}
// Return the maximum of two cases:
// (1) nth item included
// (2) not included
else {
return max( val[n-1] + knapSack(W-wt[n-1], wt, val, n-1),
knapSack(W, wt, val, n-1)
);
}
}
// Driver program to test above function
public static void main(String args[])
{
int val[] = new int[]{29,74,16,55,52,75,74,35,78};
float wt[] = new float[]{85.31f,14.55f,3.98f,26.24f,63.69f,76.25f,60.02f,93.18f,89.95f};
float W = 75f;
int n = val.length;
System.out.println(knapSack(W, wt, val, n));
}
}
当前结果：148
预期结果： 2,7
current result: 148 expected result: 2,7
推荐答案
在这里您可以执行此操作（尽管它会占用一些额外的内存）-
Here's how you can do it (though it uses some extra of memory)-
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
/* A Naive recursive implementation of 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(float W, float wt[], int val[], int n,int visited[])
{
// Base Case
if (n == 0 || W == 0)
return 0;
// If weight of the nth item is more than Knapsack capacity W, then
// this item cannot be included in the optimal solution
if (wt[n-1] > W)
{
return knapSack(W, wt, val, n-1,visited);
}
// Return the maximum of two cases:
// (1) nth item included
// (2) not included
else {
int v1[]=new int[visited.length];
System.arraycopy(visited, 0, v1, 0, v1.length);
int v2[]=new int[visited.length];
System.arraycopy(visited, 0, v2, 0, v2.length);
v1[n-1]=1;
int ans1 = val[n-1] + knapSack(W-wt[n-1], wt, val, n-1,v1);
int ans2 = knapSack(W, wt, val, n-1,v2);
if(ans1>ans2){
System.arraycopy(v1, 0, visited, 0, v1.length);
return ans1;
}
else{
System.arraycopy(v2, 0, visited, 0, v2.length);
return ans2;
}
}
}
// Driver program to test above function
public static void main(String args[])
{
int val[] = new int[]{29,74,16,55,52,75,74,35,78};
float wt[] = new float[]{85.31f,14.55f,3.98f,26.24f,63.69f,76.25f,60.02f,93.18f,89.95f};
float W = 75f;
int n = val.length;
int visited[] = new int[n];
System.out.println(knapSack(W, wt, val, n, visited));
for(int i=0;i<n;i++)
if(visited[i]==1)
System.out.println(i+1);
}
}
我所做的是我创建了一个访问数组，并且如果使用当前元素，而不是标记为已访问当前元素，则它保持为零。最后，我遍历此数组并打印访问的每个元素为1
What i did is that i created a visited array and if current element is used than i marked visited of current element one otherwise it remains zero. Finally, i iterated through this array and printed every element for which visited is 1
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