弗洛伊德算法（Floyd）

import java.util.Arrays;

public class FloydAlgorithm {

    public static void main(String[] args) {
        char[] vertexs = {‘A‘, ‘B‘, ‘C‘, ‘D‘, ‘E‘, ‘F‘, ‘G‘};
        int[][] matrix = new int[vertexs.length][vertexs.length];
        final int N = 65535;
        matrix[0] = new int[]{0, 5, 7, N, N, N, 2};
        matrix[1] = new int[]{5, 0, N, 9, N, N, 3};
        matrix[2] = new int[]{7, N, 0, N, 8, N, N};
        matrix[3] = new int[]{N, 9, N, 0, N, 4, N};
        matrix[4] = new int[]{N, N, 8, N, 0, 5, 4};
        matrix[5] = new int[]{N, N, N, 4, 5, 0, 6};
        matrix[6] = new int[]{2, 3, N, N, 4, 6, 0};
        Graph graph = new Graph(vertexs.length, matrix, vertexs);
        graph.show();
        graph.floyd();
        graph.show();
    }
}


class Graph {
    private char[] vertex;//顶点数组
    private int[][] dis;//距离表
    private int[][] pre;//前驱节点表

    /**
     * @param lenght 大小
     * @param matrix 邻接矩阵
     * @param vertex 顶点数组
     */
    public Graph(int lenght, int[][] matrix, char[] vertex) {
        this.vertex = vertex;
        this.dis = matrix;
        this.pre = new int[lenght][lenght];
        for (int i = 0; i < lenght; i++) {
            Arrays.fill(pre[i], i);
        }
    }

    //显示 pre dis
    public void show() {
        char[] vertexs = {‘A‘, ‘B‘, ‘C‘, ‘D‘, ‘E‘, ‘F‘, ‘G‘};
        for (int k = 0; k < dis.length; k++) {
            for (int i = 0; i < dis.length; i++) {
                System.out.print(vertexs[pre[k][i]] + " ");
            }
            System.out.println();
            for (int i = 0; i < dis.length; i++) {
                System.out.print("(" + vertexs[k] + "->" + vertexs[i] + " " + dis[k][i] + ") ");
            }
            System.out.println();
            System.out.println();
        }
    }

    public void floyd() {
        int len = 0;
        //对中间顶点遍历 k 是中间顶点的下标
        for (int k = 0; k < dis.length; k++) {
            for (int i = 0; i < dis.length; i++) {
                for (int j = 0; j < dis.length; j++){
                    len = dis[i][k] + dis[k][j];
                    if (len < dis[i][j]){
                        dis[i][j] = len;
                        pre[i][j] = pre[k][j];
                    }
                }
            }
        }
    }
}