遞歸蠻力迷宮求解器Java

Question

為了編寫一個解決C程序的蠻力迷宮，我首先編寫了這個Java程序來測試一個想法。 我對C還是很陌生，打算在Java中正確處理之后將其轉換。 結果，我試圖遠離arraylist，fancy庫等，以便更輕松地轉換為C。該程序需要生成最短步驟的單個寬度路徑來解決迷宮。 我認為我的問題可能在於將通過每個遞歸傳遞的路徑存儲數組分段。 感謝您的關注。 -喬

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 

代碼中解釋了數字符號

    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 + "-------");
}

}

Answer 1

這本來不是要作為答案的，但實際上已經演變成一個答案。 老實說，我認為從Java開始並轉向C是一個壞主意，因為這兩種語言確實沒有什么相似之處，而且您不會幫上忙，因為如果您依賴於Java的任何功能，在移植它時都會遇到嚴重的問題有C沒有（即大多數）

也就是說，我將草擬一些算法C的東西。

支持結構

typedef
struct Node
{
    int x, y;
    // x and y are array indices
}
Node;

typedef
struct Path
{
    int maxlen, head;
    Node * path;
    // maxlen is size of path, head is the index of the current node
    // path is the pointer to the node array
}
Path;

int    node_compare(Node * n1, Node * n2); // returns true if nodes are equal, else false

void   path_setup(Path * p, Node * n); // allocates Path.path and sets first node
void   path_embiggen(Path * p);        // use realloc to make path bigger in case it fills up
int    path_toosmall(Path * p);        // returns true if the path needs to be reallocated to add more nodes
Node * path_head(Path * p);            // returns the head node of the path
void   path_push(Path * p, Node * n);  // pushes a new head node onto the path
void   path_pop(Path * p);             // pops a node from path

您可能將迷宮格式更改為鄰接列表之類的東西。 您可以將每個節點存儲為一個掩碼，詳細說明可以從該節點移動到的節點。

迷宮格式

const int // these constants indicate which directions of travel are possible from a node
N = (1 << 0),       // travel NORTH from node is possible
S = (1 << 1),       // travel SOUTH from node is possible
E = (1 << 2),       // travel EAST  from node is possible
W = (1 << 3),       // travel WEST  from node is possible
NUM_DIRECTIONS = 4; // number of directions (might not be 4.  no reason it has to be)

const int
START  = (1 << 4),  // starting  node
FINISH = (1 << 5);  // finishing node

const int
MAZE_X = 4,         // maze dimensions
MAZE_Y = 4;

int maze[MAZE_X][MAZE_Y] = 
{
    {E,        S|E|W,    S|E|W,    S|W       },
    {S|FINISH, N|S,      N|START,  N|S       },
    {N|S,      N|E,      S|E|W,    N|S|W     },
    {N|E,      E|W,      N|W,      N         }
};

Node start  = {1, 2}; // position of start node
Node finish = {1, 0}; // position of end node

我的迷宮與您的迷宮不同：這兩種格式彼此並非完全一對一地映射。 例如，您的格式允許更精細的移動，而我的格式允許單向路徑。

請注意，您的格式明確地定位了牆。 以我的格式，牆在概念上位於不可能建立路徑的任何地方。 我創建的迷宮有3個水平牆和5個垂直牆（並且也是封閉的，即有連續的牆圍住整個迷宮）

為了進行遍歷，我將使用深度優先搜索。 您可以通過多種方式將標志映射到路線，例如以下。 由於無論如何都要遍歷每個對象，因此訪問時間是無關緊要的，因此僅使用數組而不是某種更快的關聯容器就足夠了。

數據格式到偏移映射

// map directions to array offsets
// format is [flag], [x offset], [y offset]
int mappings[][] =
{
    {N, -1,  0},
    {S,  1,  0},
    {E,  0,  1},
    {W,  0, -1}
}

最后，您的搜索。 您可以迭代或遞歸實現。 我的示例使用遞歸。

搜索算法偽代碼

int search_for_path(int ** maze, char ** visited, Path * path)
{
    Node * head = path_head(path);
    Node temp;
    int i;

    if (node_compare(head, &finish)) return 1; // found finish
    if (visited[head->x][head->y])   return 0; // don't traverse again, that's pointless

    visited[head->x][head->y] = 1;
    if (path_toosmall(path)) path_embiggen(path);

    for (i = 0; i < NUM_DIRECTIONS; ++i)
    {
        if (maze[head->x][head->y] & mappings[i][0]) // path in this direction
        {
            temp = {head->x + mappings[i][1], head->y + mappings[i][2]};
            path_push(path, &temp);
            if (search_for_path(maze, visited, path)) return 1; // something found end
            path_pop(path);
        }
    }
    return 0; // unable to find path from any unvisited neighbor
}

要調用此函數，您應該像這樣設置所有內容：

調用解算器

// we already have the maze
// int maze[MAZE_X][MAZE_Y] = {...};

// make a visited list, set to all 0 (unvisited)
int visited[MAZE_X][MAZE_Y] = 
{
    {0,0,0,0},
    {0,0,0,0},
    {0,0,0,0},
    {0,0,0,0}
};

// setup the path
Path p;
path_setup(&p, &start);

if (search_for_path(maze, visited, &path))
{
    // succeeded, path contains the list of nodes containing coordinates from start to end
}
else
{
    // maze was impossible
}

值得注意的是，因為我在編輯框中都寫了這些，所以我還沒有測試過。 第一次嘗試可能無法正常工作，可能需要花些時間。 例如，除非全局聲明了開始和結束，否則將存在一些問題。 最好將目標節點傳遞給搜索功能，而不要使用全局變量。