We can solve using kruskal's approach + union-find (disjoint set). Whenever we are trying to add edge which leads to any of the machines we need to skip adding that edge and consider it in the answer.

class Result {

/*
 * Complete the 'minTime' function below.
 *
 * The function is expected to return an INTEGER.
 * The function accepts following parameters:
 * 1. 2D_INTEGER_ARRAY roads
 * 2. INTEGER_ARRAY machines
 */
private static int  find(int x, Map<Integer,Integer> parent)
{
    int t = parent.getOrDefault(x, x);
    if(t == x) return x;
    parent.put(x, find(parent.get(x),parent));

    return parent.get(x);
}
private static int union(List<Integer> road, boolean[] red,int[] size,Map<Integer,Integer> parent){
    int root1 = find(road.get(0), parent);
    int root2 = find(road.get(1), parent);
    if(red[root1] && red[root2]) return road.get(2);

    if(root1 != root2) {
        if(size[root1] > size[root2]) {
            int r = root1;
            root1 = root2;
            root2 = r;
        }
        parent.put(root1, root2);
        size[root2] += size[root1];
    }

    red[root1] |= red[root2];
    red[root2] |= red[root1];
    return 0;
}

public static int minTime(List<List<Integer>> roads, List<Integer> machines) {
    // Write your code here
    Collections.sort(roads, (a, b) -> {
        if (a.get(2) > b.get(2))
            return -1;
        else if (a.get(2) < b.get(2))
            return 1;
        return 0;
    });
    Map<Integer,Integer> parent = new HashMap<>();

    int n = roads.size()+1;
    boolean[] red = new boolean[n];
    for(int machine: machines) red[machine]=true;

    int[] size = new int[n];
    for(int i = 0; i < n; i++) size[i] = 1;

    int totalCost = 0;

    for(List<Integer> road:roads)
    {

        totalCost += union(road,red,size,parent);
    }
  //System.out.println(Arrays.toString();
    return totalCost;

}

}

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