Question

我正在研究如何使用A *算法来查找路径，我希望看到这样做的最佳方法。这就是我想要有一个起点和终点，然后通过构建函数构建迷宫，然后用产科填充它然后在A *算法中打印出一个表格中的路线，就像格式基本上将0更改为a 3表示所采取的路径（产科将等于1）。这听起来像个好计划吗？

我遇到的麻烦是我不知道放入数组中产科的最佳方法。这就是我到目前为止所做的：

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

很抱歉，但我不明白为什么你为障碍宣布了这么多的二维数组...... 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你想要的意思是（220,616），（220,666），（251,670），（272,647）是1个障碍的角点。

如果是这样，那么我建议不要采用4个二维数组，而是用迷宫[] []数组中的障碍物覆盖无穷大，即最高整数no。（让我们把它看作10000）

和其他（x，y）位置，在迷宫[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 *算法从头到尾进行覆盖。

迷宫与路径寻找算法

1 个答案: