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Java版-图论-最小生成树-Kruskal算法

news 2025/7/19 6:24:39

实现描述

为了造出一棵最小生成树，我们从最小边权的边开始，按边权从小到大依次加入，如果某次加边产生了环，就扔掉这条边，直到加入了 n-1 条边，即形成了一棵树。

实现代码

首选我们对所有的边，按照权重排序；
之后，从小到大选择边，如果当前的边已经连通过了，则放弃此边，查看下一条边；若没有连通过，通过并查集进行连通；
直至所有点都访问过，此时，完成。

在这里插入图片描述

如上图，节点0~5，边关系如上；
先对边权重进行从小到大排序，得到：

[{"u":4,"v":5,"weight":1
},{"u":0,"v":5,"weight":3
},{"u":1,"v":2,"weight":4
},{"u":0,"v":1,"weight":5
},{"u":2,"v":3,"weight":5
},{"u":3,"v":4,"weight":5
},{"u":3,"v":5,"weight":6
},{"u":0,"v":3,"weight":7
},{"u":0,"v":2,"weight":8
},{"u":2,"v":5,"weight":9
}]

然后依次选择排序后的边：
在这里插入图片描述
下面代码对应上图数据及其过程：

import com.alibaba.fastjson.JSONObject;
import lombok.AllArgsConstructor;
import lombok.Data;
import lombok.NoArgsConstructor;
import org.jetbrains.annotations.NotNull;import java.util.*;public class kruskal {/*** 边定义*/@Data@AllArgsConstructor@NoArgsConstructorstatic class Edge implements Comparable<Edge> {int u, v;int weight;@Overridepublic int compareTo(@NotNull Edge o) {return weight - o.weight;}}static List<Edge> getKruskalEdges(int nodeNum, int[][] grid) {List<Edge> result = new LinkedList<>();List<Edge> edges = new LinkedList<>();for (int i = 0; i < grid.length; i++) {int u = grid[i][0];int v = grid[i][1];int weight = grid[i][2];edges.add(new Edge(u, v, weight));}Collections.sort(edges);UnionFindTemplate uf = new UnionFindTemplate(nodeNum);Set<Integer> visited = new HashSet<>();for (Edge edge : edges) {if (uf.connected(edge.u, edge.v)) {continue;}uf.union(edge.u, edge.v);visited.add(edge.u);visited.add(edge.v);result.add(edge);if (visited.size() == nodeNum) {break;}}return result;}public static void main(String[] args) {int nodeNum = 6;int[][] grid = {{0, 1, 5},{0, 5, 3},{0, 3, 7},{0, 2, 8},{1, 2, 4},{2, 5, 9},{3, 5, 6},{2, 3, 5},{3, 4, 5},{4, 5, 1}};System.out.println(JSONObject.toJSONString(getKruskalEdges(nodeNum, grid)));}
}

其中，并查集模版的实现如下：

public class UnionFindTemplate {int[] parent;int[] size;int n;public int setCount;//连通分量个数public UnionFindTemplate(int n) {this.n = n;this.parent = new int[n];this.size = new int[n];setCount = n;Arrays.fill(this.size, 1);for (int i = 0; i < n; ++i) {parent[i] = i;}}public int findParent(int x) {if (parent[x] == x) {return x;} else {parent[x] = findParent(parent[x]);return parent[x];}}public void union(int x, int y) {x = findParent(x);y = findParent(y);if (x == y) {return;}if (size[x] < size[y]) {int temp = x;x = y;y = temp;}//y合并到xparent[y] = x;size[x] += size[y];setCount--;}public boolean connected(int x, int y) {x = findParent(x);y = findParent(y);return x == y;}}