パスを見つけるために A* アルゴリズムを使用する方法の学習に取り組んでおり、これを行う最善の方法を見たいと思っています。これは私が考えていたものです。開始点と終了点を設定し、ビルド関数を介して迷路を作成し、産科を入力してから、A * アルゴリズムを実行し、基本的に 0 を 0 に変更するような形式でルートを印刷します。 3 は、たどられたパスを示します (産科は 1 に等しくなります)。これは良い計画のように聞こえますか?
私が問題を抱えているのは、産科を配列に入れる最良の方法がわからないということです。これは私がこれまでに持っているものです:
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()
{
}
}
これを手伝ってくれてありがとう