Question

我有一个图G（V，E），如果存在一个图，我必须计算一个包含 e （边缘属于E）的MST（最小生成树）。我以为我可以使用Kruskal算法在列表顶部插入弧 e ，以使算法首先选择该弧并构建包含 e 的树。这是正确的推理方法吗？如果不存在包含 e 的MST，弧会被忽略吗？

这是我要更改的Java代码：

import java.util.ArrayList;
import java.util.Comparator;
import java.util.PriorityQueue;

public class KrushkalMST {
    static class Edge {
        int source;
        int destination;
        int weight;

        public Edge(int source, int destination, int weight) {
            this.source = source;
            this.destination = destination;
            this.weight = weight;
        }
    }

    static class Graph {
        int vertices;
        ArrayList<Edge> allEdges = new ArrayList<>();

        Graph(int vertices) {
            this.vertices = vertices;
        }

        public void addEgde(int source, int destination, int weight) {
            Edge edge = new Edge(source, destination, weight);
            allEdges.add(edge); //add to total edges
        }

        public void kruskalMST(){
            PriorityQueue<Edge> pq = new PriorityQueue<>(allEdges.size(), Comparator.comparingInt(o -> o.weight));

            //add all the edges to priority queue, //sort the edges on weights
            for (int i = 0; i <allEdges.size() ; i++) {
                pq.add(allEdges.get(i));
            }

            //create a parent []
            int [] parent = new int[vertices];

            //makeset
            makeSet(parent);

            ArrayList<Edge> mst = new ArrayList<>();

            //process vertices - 1 edges
            int index = 0;
            while(index<vertices-1){
                Edge edge = pq.remove();
                //check if adding this edge creates a cycle
                int x_set = find(parent, edge.source);
                int y_set = find(parent, edge.destination);

                if(x_set==y_set){
                    //ignore, will create cycle
                }else {
                    //add it to our final result
                    mst.add(edge);
                    index++;
                    union(parent,x_set,y_set);
                }
            }
            //print MST
            System.out.println("Minimum Spanning Tree: ");
            printGraph(mst);
        }

        public void makeSet(int [] parent){
            //Make set- creating a new element with a parent pointer to itself.
            for (int i = 0; i <vertices ; i++) {
                parent[i] = i;
            }
        }

        public int find(int [] parent, int vertex){
            //chain of parent pointers from x upwards through the tree
            // until an element is reached whose parent is itself
            if(parent[vertex]!=vertex)
                return find(parent, parent[vertex]);;
            return vertex;
        }

        public void union(int [] parent, int x, int y){
            int x_set_parent = find(parent, x);
            int y_set_parent = find(parent, y);
            //make x as parent of y
            parent[y_set_parent] = x_set_parent;
        }

        public void printGraph(ArrayList<Edge> edgeList){
            for (int i = 0; i <edgeList.size() ; i++) {
                Edge edge = edgeList.get(i);
                System.out.println("Edge-" + i + " source: " + edge.source +
                        " destination: " + edge.destination +
                        " weight: " + edge.weight);
            }
        }
    }
    public static void main(String[] args) {
            int vertices = 6;
            Graph graph = new Graph(vertices);
            graph.addEgde(0, 1, 4);
            graph.addEgde(0, 2, 3);
            graph.addEgde(1, 2, 1);
            graph.addEgde(1, 3, 2);
            graph.addEgde(2, 3, 4);
            graph.addEgde(3, 4, 2);
            graph.addEgde(4, 5, 6);
            graph.kruskalMST();
    }
}

Answer 1

如果您打算实现Kruskal算法，则必须实现方法union和find。这些方法正在创建树结构。首先，将每个节点初始化为其树的根，然后，如果不创建循环，则使用并集将它们合并。

因此，要回答您的问题，只需在运行算法之前通过联合查找创建所需的集合即可。

KRUSKAL最小跨树

1 个答案: