ブルート フォースの迷路を解く C プログラムを作成する試みとして、最初にこの Java プログラムを作成してアイデアをテストしました。私はCに非常に慣れていないので、Javaでこれを正しく行った後に変換するつもりです。その結果、C への変換を容易にするために、配列リストやファンシー ライブラリなどから離れようとしています。プログラムは、迷路を解くための最短ステップの単一幅パスを生成する必要があります。私の問題は、各再帰を通過したパス格納配列を断片化することにあると思います。これを見てくれてありがとう。-ジョー
maze:
1 3 3 3 3
3 3 3 3 3
3 0 0 0 3
3 0 3 3 3
0 3 3 3 2
Same maze solved by this program:
4 4 4 4 4
4 4 4 4 4
4 0 0 0 4
3 0 3 3 4
0 3 3 3 2
数値表記はコードで説明されています
public class javamaze {
static storage[] best_path;
static int best_count;
static storage[] path;
//the maze - 1 = start; 2 = finish; 3 = open path
static int maze[][] = {{1, 3, 3, 3, 3},
{3, 3, 3, 3, 3},
{0, 0, 0, 0, 3},
{0, 0, 3, 3, 3},
{3, 3, 3, 3, 2}};
public static void main(String[] args) {
int count1;
int count2;
//declares variables used in the solve method
best_count = 0;
storage[] path = new storage[10000];
best_path = new storage[10000];
int path_count = 0;
System.out.println("Here is the maze:");
for(count1 = 0; count1 < 5; count1++) {
for(count2 = 0; count2 < 5; count2++) {
System.out.print(maze[count1][count2] + " ");
}
System.out.println("");
}
//solves the maze
solve(findStart()/5, findStart()%5, path, path_count);
//assigns an int 4 path to the maze to visually represent the shortest path
for(int count = 0; count <= best_path.length - 1; count++)
if (best_path[count] != null)
maze[best_path[count].getx()][best_path[count].gety()] = 4;
System.out.print("Here is the solved maze\n");
//prints the solved maze
for(count1 = 0; count1 < 5; count1++) {
for(count2 = 0; count2 < 5; count2++){
System.out.print(maze[count1][count2] + " ");
}
System.out.print("\n");
}
}
//finds maze start marked by int 1 - this works perfectly and isn't related to the problem
public static int findStart() {
int count1, count2;
for(count1 = 0; count1 < 5; count1++) {
for(count2 = 0; count2 < 5; count2++) {
if (maze[count1][count2] == 1)
return (count1 * 5 + count2);
}
}
return -1;
}
//saves path coordinate values into a new array
public static void save_storage(storage[] old_storage) {
int count;
for(count = 0; count < old_storage.length; count++) {
best_path[count] = old_storage[count];
}
}
//solves the maze
public static Boolean solve(int x, int y, storage[] path, int path_count) {
//checks to see if grid squares are valid (3 = open path; 0 = wall
if (x < 0 || x > 4) { //array grid is a 5 by 5
//System.out.println("found row end returning false");
return false;
}
if (y < 0 || y > 4) {
//System.out.println("Found col end returning false");
return false;
}
//when finding finish - records the number of moves in static int best_count
if (maze[x][y] == 2) {
if (best_count == 0 || best_count > path_count) {
System.out.println("Found end with this many moves: " + path_count);
best_count = path_count;
save_storage(path); //copies path counting array into a new static array
}
}
//returns false if it hits a wall
if (maze[x][y] == 0)
return false;
//checks with previously crossed paths to prevent an unnecessary repeat in steps
for(storage i: path)
if (i != null)
if (i.getx() == x && i.gety() == y)
return false;
//saves current recursive x, y (row, col) coordinates into a storage object which is then added to an array.
//this array is supposed to fragment per each recursion which doesn't seem to - this may be the issue
storage storespoints = new storage(x, y);
path[path_count] = storespoints;
//recurses up, down, right, left
if (solve((x-1), y, path, path_count++) == true || solve((x+1), y, path, path_count++) == true ||
solve(x, (y+1), path, path_count++) == true || solve(x, (y-1), path, path_count++) == true) {
return true;
}
return false;
}
}
//stores (x, y) aka row, col coordinate points
class storage {
private int x;
private int y;
public storage(int x, int y) {
this.x = x;
this.y = y;
}
public int getx() {
return x;
}
public int gety() {
return y;
}
public String toString() {
return ("storage coordinate: " + x + ", " + y + "-------");
}
}