Java实现最小生成树MST的两种解法

package Prim;
 
import java.util.Arrays;
 
public class PrimAlGorithm {
    public static void main(String[] args) {
        //测试看看图是否创建ok
        char[] data = new char[]{'A', 'B', 'C', 'D', 'E', 'F', 'G'};
        int verxs = data.length;
        //邻接矩阵的关系使用二维数组表示,10000这个大数，表示两个点不联通
        int[][] weight = new int[][]{
                {10000, 5, 7, 10000, 10000, 10000, 2},
                {5, 10000, 10000, 9, 10000, 10000, 3},
                {7, 10000, 10000, 10000, 8, 10000, 10000},
                {10000, 9, 10000, 10000, 10000, 4, 10000},
                {10000, 10000, 8, 10000, 10000, 5, 4},
                {10000, 10000, 10000, 4, 5, 10000, 6},
                {2, 3, 10000, 10000, 4, 6, 10000},};
        MGraph mGraph = new MGraph(verxs);
        MinTree minTree = new MinTree();
        minTree.createGraph(mGraph, verxs, data, weight);
        minTree.showGraph(mGraph);
        minTree.Prim(mGraph, 0);
    }
}
 
class MinTree {
    
    public void createGraph(MGraph graph, int verxs, char[] data, int[][] weight) {
        int i, j;
        for (i = 0; i < verxs; i++) {
            graph.data[i] = data[i];
            for (j = 0; j < verxs; j++) {
                graph.weight[i][j] = weight[i][j];
            }
        }
    }
 
    // 显示图的邻接矩阵
    public void showGraph(MGraph graph) {
        for (int[] link : graph.weight) {
            System.out.println(Arrays.toString(link));
        }
    }
 
    
    public void Prim(MGraph graph, int v) {
        //定义一个数组，判断节点是不是被访问过了
        int[] visited = new int[graph.verxs];
        //v这个点已经被访问了，从这个点开始访问
        visited[v] = 1;
        //找到节点下标
        int h1 = -1;
        int h2 = -1;
        int minWeight = 10000;//定义初始值为最大值，只要出现小的就会替换
        int sum = 0;
        // 从1开始循环，相当于就是生成graph.verx - 1条边
        for (int k = 1; k < graph.verxs; k++) {
 
            for (int i = 0; i < graph.verxs; i++) {//遍历已经访问过的点
                if (visited[i] == 1){
                    for (int j = 0; j < graph.verxs; j++) {//遍历没有访问过的点
                        //在未访问点中寻找所有与访问过的点相连的边中权值最小值
                        if (visited[i] == 1 && visited[j] == 0 && graph.weight[i][j] < minWeight) {
                            minWeight = graph.weight[i][j];
                            h1 = i;
                            h2 = j;
                        }
                    }
                }
            }
            sum += minWeight; // 求最小生成熟的总权值
            //此时已经找到一条边是最小了
            System.out.println("边<" + graph.data[h1] + "," + graph.data[h2] + "> 权值:" + minWeight);
            //然后标记点
            visited[h2] = 1;
            //将权值重新变成最大值
            minWeight = 10000;
        }
        System.out.println("最小生成树的权值是:" + sum);
 
    }
}
 
// 图
class MGraph {
    int verxs; // 表示图节点个数
    char[] data; // 表示节点数据
    int[][] weight; // 表示边
 
    public MGraph(int verxs) {
        this.verxs = verxs;
        data = new char[verxs];
        weight = new int[verxs][verxs];
    }
}

package Kruskal;
 
import java.util.Arrays;
 
public class KruskalCase {
 
    private int edgeNum; //边的个数
    private char[] vertexs; //顶点数组
    private int[][] matrix; //邻接矩阵
    //使用 INF 表示两个顶点不能连通
    private static final int INF = Integer.MAX_VALUE;
 
    public static void main(String[] args) {
        char[] vertexs = {'A', 'B', 'C', 'D', 'E', 'F', 'G'};
        //克鲁斯卡尔算法的邻接矩阵
        int matrix[][] = {
                
                 {0, 12, INF, INF, INF, 16, 14},
                 {12, 0, 10, INF, INF, 7, INF},
                 {INF, 10, 0, 3, 5, 6, INF},
                 {INF, INF, 3, 0, 4, INF, INF},
                 {INF, INF, 5, 4, 0, 2, 8},
                 {16, 7, 6, INF, 2, 0, 9},
                 {14, INF, INF, INF, 8, 9, 0}};
        //大家可以在去测试其它的邻接矩阵，结果都可以得到最小生成树.
 
        //创建KruskalCase 对象实例
        KruskalCase kruskalCase = new KruskalCase(vertexs, matrix);
        //输出构建的
        kruskalCase.print();
        kruskalCase.kruskal();
 
    }
 
    //构造器
    public KruskalCase(char[] vertexs, int[][] matrix) {
        //初始化顶点数和边的个数
        int vlen = vertexs.length;
 
        //初始化顶点, 复制拷贝的方式
        this.vertexs = new char[vlen];
        for (int i = 0; i < vertexs.length; i++) {
            this.vertexs[i] = vertexs[i];
        }
 
        //初始化边, 使用的是复制拷贝的方式
        this.matrix = new int[vlen][vlen];
        for (int i = 0; i < vlen; i++) {
            for (int j = 0; j < vlen; j++) {
                this.matrix[i][j] = matrix[i][j];
            }
        }
        //统计边的条数
        for (int i = 0; i < vlen; i++) {
            for (int j = i + 1; j < vlen; j++) {
                if (this.matrix[i][j] != INF) {
                    edgeNum++;
                }
            }
        }
 
    }
 
    public void kruskal() {
        int index = 0; //表示最后结果数组的索引
        int[] ends = new int[edgeNum]; //用于保存"已有最小生成树" 中的每个顶点在最小生成树中的终点
        //创建结果数组, 保存最后的最小生成树
        EData[] rets = new EData[edgeNum];
 
        //获取图中 所有的边的集合 ， 一共有12边
        EData[] edges = getEdges();
        System.out.println("图的边的集合=" + Arrays.toString(edges) + " 共" + edges.length); //12
 
        //按照边的权值大小进行排序(从小到大)
        sortEdges(edges);
 
        //遍历edges 数组，将边添加到最小生成树中时，判断是准备加入的边否形成了回路，如果没有，就加入 rets, 否则不能加入
        for (int i = 0; i < edgeNum; i++) {
            //获取到第i条边的第一个顶点(起点)
            int p1 = getPosition(edges[i].start); //p1=4
            //获取到第i条边的第2个顶点
            int p2 = getPosition(edges[i].end); //p2 = 5
 
            //获取p1这个顶点在已有最小生成树中的终点
            int m = getEnd(ends, p1); //m = 4
            //获取p2这个顶点在已有最小生成树中的终点
            int n = getEnd(ends, p2); // n = 5
            //是否构成回路
            if (m != n) { //没有构成回路
                ends[m] = n; // 设置m 在"已有最小生成树"中的终点 <E,F> [0,0,0,0,5,0,0,0,0,0,0,0]
                rets[index++] = edges[i]; //有一条边加入到rets数组
            }
        }
        //<E,F> <C,D> <D,E> <B,F> <E,G> <A,B>。
        //统计并打印 "最小生成树", 输出  rets
        System.out.println("最小生成树为");
        for (int i = 0; i < index; i++) {
            System.out.println(rets[i]);
        }
 
 
    }
 
    //打印邻接矩阵
    public void print() {
        System.out.println("邻接矩阵为: \n");
        for (int i = 0; i < vertexs.length; i++) {
            for (int j = 0; j < vertexs.length; j++) {
                System.out.printf("%12d", matrix[i][j]);
            }
            System.out.println();//换行
        }
    }
 
    
    private void sortEdges(EData[] edges) {
        for (int i = 0; i < edges.length - 1; i++) {
            for (int j = 0; j < edges.length - 1 - i; j++) {
                if (edges[j].weight > edges[j + 1].weight) {//交换
                    EData tmp = edges[j];
                    edges[j] = edges[j + 1];
                    edges[j + 1] = tmp;
                }
            }
        }
    }
 
    
    private int getPosition(char ch) {
        for (int i = 0; i < vertexs.length; i++) {
            if (vertexs[i] == ch) {//找到
                return i;
            }
        }
        //找不到,返回-1
        return -1;
    }
 
    
    private EData[] getEdges() {
        int index = 0;
        EData[] edges = new EData[edgeNum];
        for (int i = 0; i < vertexs.length; i++) {
            for (int j = i + 1; j < vertexs.length; j++) {
                if (matrix[i][j] != INF) {
                    edges[index++] = new EData(vertexs[i], vertexs[j], matrix[i][j]);
                }
            }
        }
        return edges;
    }
 
    
    private int getEnd(int[] ends, int i) { // i = 4 [0,0,0,0,5,0,0,0,0,0,0,0]
        while (ends[i] != 0) {
            i = ends[i];
        }
        return i;
    }
 
}
 
//创建一个类EData ，它的对象实例就表示一条边
class EData {
    char start; //边的一个点
    char end; //边的另外一个点
    int weight; //边的权值
 
    //构造器
    public EData(char start, char end, int weight) {
        this.start = start;
        this.end = end;
        this.weight = weight;
    }
 
    //重写toString, 便于输出边信息
    @Override
    public String toString() {
        return "EData [<" + start + ", " + end + ">= " + weight + "]";
    }
 
 
}