Puli Mu algorithm base

Puli Mu algorithm

Scenarios - road problem

Minimum spanning tree

The nature of the problem road is the minimum spanning tree problem, first introduce minimum spanning tree (Minimum Cost Spanning Tree), referred to as MST

1. Given a non-weighted FIG communicated, how to select a spanning tree, so thatThe minimum sum of all the trees on the edge of the rightThis is called the minimum spanning tree
2. N vertices, N-1 must have edges
3. It contains all the vertices
4. N-1 edges in FIG.
5. Minimum spanning tree algorithm is mainlyPuli Mu algorithmwithKruskal algorithm

Algorithms Introduction

1. Puli Mu minimum spanning tree algorithm, which is in communication with FIG comprising n vertices, find only the (n-1) edges link subgraph comprising of all n vertices, the so-calledMinimal connected subgraph

2. Puli Mu algorithm is as follows:

   

3. Road illustration question:

Code

package L十大算法.Prime;


import java.util.Arrays;

/**
 * @Author Zhou  jian
 * @Date 2020 ${month}  2020/1/15 0015  15:43
 * 普利姆算法
 */
public class PrimeAlgrothim {

    public static void main(String[] args) {

        //测试看看图是否创建成功
        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},};

        //创建Graph对象
        MGraph mgraph = new MGraph(verxs);

        //创建一个MiniTree
        MinTree minTree = new MinTree();
        minTree.createGraph(mgraph,verxs,data,weight);

        //输出
        minTree.showGraph(mgraph);

        //测试普利姆水泥发
        minTree.prime(mgraph,0);
    }
}


//最小生成树---->村庄的图
class MinTree{

    /**
     *
     * @param graph   图对象
     * @param verxs   图对应的顶点个数
     * @param data   图的各个顶点的值
     * @param weight  图的临街矩阵
     */
    //创建图的领结矩阵
    public  void createGraph(MGraph graph,int verxs,char[] data,int[][] weight){
        int i,j;
        for(i=0;i<verxs;i++){  //顶点
            graph.data[i] = data[i];
            for(j=0;j<verxs;j++){
                graph.weight[i][j] = weight[i][j];
            }
        }
    }
        //显示图的临街矩阵
        public void showGraph(MGraph graph){
            for(int i=0;i<graph.verxs;i++){
                for(int[] link : graph.weight){
                    System.out.println(Arrays.toString(link));
                }
            }
         }



         //编写prime算法，得到最小生成树
        /*
         * @param graph  图
         * @param v  表示从图的第几个顶点开始生成'A'->0
         */
        public void prime(MGraph graph,int v){

            //visited[]  标记节点（顶点）是否访问过
            int[] visited = new int[graph.verxs];

            //visited[] 默认元素值都是0，表示没有访问过
            for(int i=0;i<graph.verxs;i++){
                visited[i]=0;
            }

            //把当前节点标记为已经访问
            visited[v]=1;

            //h1和h2记录这两个顶点下标
            int h1 = -1;
            int h2 = -1;
            //将minWeight初始化为最大值，后面再遍历过程中，会被替换
            int minWeight = 10000;
            for(int k=1;k<graph.verxs;k++){ //因为有graph.verxs顶点，普利斯木算法结束后，有graph.versx-1边

                //这个是确定每一次生成的子图和哪个节点的距离最近
                for(int i=0;i<graph.verxs;i++){//遍历已经访问的节点；i节点表示被访问过的结案
                    for(int j=0;j<graph.verxs;j++){ //遍历没有访问的节点；j表示还没有被访问
                        if(visited[i]==1&&visited[j]==0&&graph.weight[i][j]<minWeight){
                            //替换minWeight（寻找已经访问过的节点和未访问过的节点间的权值最小的边)
                            minWeight = graph.weight[i][j];
                            h1 = i;
                            h2 = j;
                        }
                    }
                }
                //找到一条边最小
                System.out.println("边<"+graph.data[h1]+","+graph.data[h2]+">权值"+minWeight);
                //将当前节点标记为已经访问
                visited[h2]=1;
                //minWeight重新设置为最大值10000
                minWeight=10000;
            }

        }
}



class MGraph{

    int verxs; //表示图的节点的个数

    char[] data;//存放节点数据

    int[][] weight; //存放边，就是我们的邻近矩阵

    public MGraph(int verxs) {
        this.verxs = verxs;
        data = new char[verxs];
        weight = new int[verxs][verxs];
    }
}