具有寻路算法的迷宫

Question

我正在努力学习如何使用 A* 算法来寻找路径，我想看看最好的方法。这就是我在想的我想要一个起点和终点然后通过构建函数构建迷宫然后用产科填充它然后运行 ​​A* 算法在表格中打印路线基本上将 0 更改为a 3 显示所采用的路径（产科将等于 1）。这听起来是个好计划吗？

我遇到的问题是我不知道将产科放入阵列的最佳方式。 这是我目前所拥有的：

public class Maze {

/**
 * @param args the command line arguments
 */
public static void main(String[] args) {


    //start and end points in the array
    int startx = 115;
    int starty = 655;
    int endx = 380;
    int endy = 560;
    //number of collums and rows
    int row = 700;
    int col = 500;
    //size of maze
    int maze [][] = new int [row][col];

    makeMaze(row, col, maze);
    printMaze(row, col, maze);



}

 //fill mazz with 0
    private static void makeMaze(int row, int col,  int maze[][])
    {   
         //fill maze with 0 for initilization
        for(int i = 0; i < row; i++)
        {
            for(int j = 0; j < col; j++)
            {
                maze[i][j] = 0;
            }

        }
    }
    //print out array/maze
    private static void printMaze(int row, int col,  int maze[][])
    {
         //... Print array in rectangular form
        for(int i = 0; i < row; i++)
        {
            for(int j = 0; j < col; j++)
            {
                System.out.print(" " + maze[i][j] );
            }
            System.out.println("");
        }

    }
    //fill the array with obsticals
    private void makeObsticals()
    {
        //obstical 1
        //this represent the corners of the object
        int ob1Point1 [][] = new int [220][616];
        int ob1Point2 [][] = new int [220][666];
        int ob1Point3 [][] = new int [251][670];
        int ob1Point4 [][] = new int [272][647];

        //object 2
        int ob2Point1 [][] = new int [341][655];
        int ob2Point2 [][] = new int [359][667];
        int ob2Point3 [][] = new int [374][651];
        int ob2Point4 [][] = new int [366][577];

        //obejct 3
        int ob3Point1 [][] = new int [311][530];
        int ob3Point2 [][] = new int [311][559];
        int ob3Point3 [][] = new int [339][578];
        int ob3Point4 [][] = new int [361][560];
        int ob3Point5 [][] = new int [361][528];
        int ob3Point6 [][] = new int [113][516];

         //object 4
        int ob4Point1 [][] = new int [105][628];
        int ob4Point2 [][] = new int [151][670];
        int ob4Point3 [][] = new int [180][629];
        int ob4Point4 [][] = new int [156][577];
        int ob4Point5 [][] = new int [113][587];

        //object 5
        int ob5Point1 [][] = new int [118][517];
        int ob5Point2 [][] = new int [245][517];
        int ob5Point3 [][] = new int [245][577];
        int ob5Point4 [][] = new int [118][577];

        //object 6
        int ob6Point1 [][] = new int [280][583];
        int ob6Point2 [][] = new int [333][583];
        int ob6Point3 [][] = new int [333][665];
        int ob6Point4 [][] = new int [280][665];

          //object 7
        int ob7Point1 [][] = new int [252][594];
        int ob7Point2 [][] = new int [290][562];
        int ob7Point3 [][] = new int [264][538];

          //object 8
        int ob8Point1 [][] = new int [198][635];
        int ob8Point2 [][] = new int [217][574];
        int ob8Point3 [][] = new int [182][574];


    }
    //astar algorithum
    private void findPath()
    {
    }

}

感谢您对此的任何帮助

Answer 1

对不起，我不明白你为什么要声明这么多二维数组作为障碍物...... 如你所说。 . .

//obstacle1
//this represent the corners of the object
   int ob1Point1 [][] = new int [220][616];
   int ob1Point2 [][] = new int [220][666];
   int ob1Point3 [][] = new int [251][670];
   int ob1Point4 [][] = new int [272][647];

我认为从上面的代码中你想要的意思是 (220,616),(220,666),(251,670),(272,647) 是 1 个障碍物的角点。

如果是这样，那么我建议不要采用 4 个二维数组，而是将 maze[][] 数组中障碍物覆盖的区域标记为 infinity 即最高整数编号。 （让我们将其视为 10000）

对于其他 (x,y) 位置，在 maze[x][y] 中放置每个位置的启发式值（这意味着从该 (x,y) 位置到达目的地 (endx,endy) 的成本）

然后应用A*算法从头到尾到达。