高龄白菜java学习第109天（java数据结构和算法（27））

代码实现：

package Algorithm.Floyd;

public class Floyd {
    public static void main(String[] args) {
        char[] vertex = {'A', 'B', 'C', 'D', 'E', 'F', 'G'};
        int[][] matrix = new int[vertex.length][vertex.length];
        final int N = 65535;//表示不可链接

        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(vertex.length, matrix, vertex);
//        graph.show();
        graph.floyd();
        graph.show();
    }
}

class Graph {
    char[] vertex;//存放顶点的数组
    int[][] pre;//保存到达目标顶点的前驱顶点
    int[][] dis;//保存从各个顶点出发，到其他顶点的距离

    public Graph(int length, int[][] matrix, char[] vertex) {//matrix是初始邻接矩阵，也是初始的dis
        this.vertex = vertex;
        this.dis = matrix;
        this.pre = new int[length][length];
        for (int i = 0; i < length; i++) {
            for (int j = 0; j < length; j++) {
                pre[i][j] = i;//存放前驱顶点的下标
            }
        }
    }

    public void show() {
        for (int i = 0; i < dis.length; i++) {
            for (int j = 0; j < dis.length; j++) {
                System.out.print(vertex[pre[i][j]]+"  ");
            }
            System.out.println();
            for (int j = 0; j < dis.length; j++) {
                System.out.print(dis[i][j]+"  ");
            }
            System.out.println();
        }
    }

    public void floyd(){
        int len = 0;//变量保存距离
        //对中间顶点的遍历
        for (int k = 0; k < dis.length; k++) {
            //起始顶点
            for (int i = 0; i < dis.length; i++) {
                //目标顶点
                for (int j = 0; j < dis.length; j++) {
                    len = dis[i][k]+dis[k][j];
                    if (len 
十、骑士周游问题 
 

 其他思路：不在同行同列同对角线的点
 策略优化：选择下一步的下一步选择越少的，效率越高
 代码实现：
 
package Algorithm.Horse;

import java.awt.*;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;

public class Horse {
    public static int X;//棋盘的列数
    public static int Y;//棋盘的行数
    public static boolean[] visited;
    public static boolean finished;//判断棋盘是否所有位置都被访问成功

    public static void main(String[] args) {
        Y = 8;
        X = 8;
        int row = 1;
        int column = 1;
        int[][] board = new int[Y][X];
        visited = new boolean[Y * X];
        travel(board, row - 1, column - 1, 1);

        for (int[] ints : board) {
            System.out.println(Arrays.toString(ints));
        }
    }

    //需要棋盘、马所在的行和列、当前是第几步
    public static void travel(int[][] board, int row, int column, int step) {
        board[row][column] = step;
        visited[row * X + column] = true;//棋盘上的第多少个格子
        //获取当前位置能走的下一个位置的集合
        ArrayList ps = next(new Point(column, row));
        sort(ps);
        //核心：递归回溯
        while (!ps.isEmpty()) {
            Point p = ps.remove(0);
            if (!visited[p.y * X + p.x]) {
                travel(board, p.y, p.x, step + 1);
            }
        }
        if (step < X * Y && !finished) {//任务没有完成，棋盘需要置0(在回溯过程中把每个点都置0了)
            //step next(Point curPoint) {
        ArrayList ps = new ArrayList<>();//存储可走的点
        Point p1 = new Point();
        //代表左上角的点是可以走的（棋盘左上角是坐标原点）
        //5
        if ((p1.x = curPoint.x - 2) >= 0 && (p1.y = curPoint.y - 1) >= 0) {
            ps.add(new Point(p1));
        }
        //6
        if ((p1.x = curPoint.x - 1) >= 0 && (p1.y = curPoint.y - 2) >= 0) {
            ps.add(new Point(p1));
        }
        //7
        if ((p1.x = curPoint.x + 1) < X && (p1.y = curPoint.y - 2) >= 0) {
            ps.add(new Point(p1));
        }
        //0
        if ((p1.x = curPoint.x + 2) < X && (p1.y = curPoint.y - 1) >= 0) {
            ps.add(new Point(p1));
        }
        //1
        if ((p1.x = curPoint.x + 2) < X && (p1.y = curPoint.y + 1) < Y) {
            ps.add(new Point(p1));
        }
        //2
        if ((p1.x = curPoint.x + 1) < X && (p1.y = curPoint.y + 2) < Y) {
            ps.add(new Point(p1));
        }
        //3
        if ((p1.x = curPoint.x - 1) >= 0 && (p1.y = curPoint.y + 2) < Y) {
            ps.add(new Point(p1));
        }
        //4
        if ((p1.x = curPoint.x - 2) >= 0 && (p1.y = curPoint.y + 1) < Y) {
            ps.add(new Point(p1));
        }
        return ps;
    }

    //根据当前的下一步的选择位置进行非递减排序,减少回溯的可能
    public static void sort(ArrayList ps){
        ps.sort(new Comparator() {
            @Override
            public int compare(Point o1, Point o2) {
                //获取到o1点的下一步的所有位置个数
//                int count1 = next(o1).size();
//                int count2 = next(o2).size();
//                if (count1					
										


					

						
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