java - 如何使用不相交集实现迷宫？

Question

这是我使用的 DisjointSet 类：

public class DisjointSet{
    public DisjointSet(int size){
        s = new int[size];
        for(int i = 0; i < size; ++i){
            s[i] = -1;
        }
    }

    public void union(int el1, int el2){
        int root1 = find(el1);
        int root2 = find(el2);
        if(root1 == root2){
            return;
        }

        if(s[root2] < s[root1]){
            s[root1] = root2;
        }
        else{
            if(s[root1] == s[root2]){
                --s[root1];
            }
            s[root2] = root1;
        }
    }

    public int find(int x){
        if(s[x] < 0){
            return x;
        }
        else{
            s[x] = find(s[x]);
            return s[x];
        }
    }

    private int[] s;
}

这是我的迷宫课：

public class Maze
{
    public Vector<Wall> maze;
    public Vector<Room> graph;
    public Vector<Integer> path;

    public int LASTROOM;
    public int height;
    public int width;
    public Random generator;
    public DisjointSet ds;

    public int dsSize;

    public Maze(int w, int h, int seed)
    {
        width = w;
        height = h;
        dsSize = width * height;

        LASTROOM = dsSize-1;

        // Maze initialization with all walls
        maze = new Vector<Wall>();
        for(int i = 0; i < height; ++i)
        {
            for(int j = 0; j < width; ++j)
            {
                if(i > 0)
                    maze.add(new Wall(j+i*height, j+(i-1)*height));

                if(j > 0)
                    maze.add(new Wall(j+i*height, j-1+i*height));
            }
        }

        // Creation of the graph topology of the maze
        graph = new Vector<Room>();
        for(int i = 0; i < dsSize; ++i)
            graph.add(new Room(i));

        // Sort the walls of random way
        generator = new Random(seed);

        for(int i = 0; i < maze.size(); ++i)
        {
            //TODO
        }

        // Initialization of related structures
        ds = new DisjointSet(dsSize);
        path = new Vector<Integer>();
    }

    public void generate()
    {
        //TODO
    }

    public void solve()
    {
        //TODO
    }
}

很长一段时间以来，我一直在寻找一种方法来实现generate()和solve()以及对迷宫墙壁的随机排序，我似乎无法在互联网上找到任何算法或实现来做到这一点.

如果由墙连接的两个部分（房间）不在同一个集合中，则generate()应该穿过迷宫变量中的置换墙并销毁它。该方法还应该在房间图中添加一条边（每个房间都有一个名为路径的邻接列表，并且Room类有一个变量id来标识每个图的顶点）。

solve()应该解决迷宫路径并生成Maze类的向量，其中包含到达出口的房间顺序。第一个房间位于 0 处，最后一个房间位于LASTROOM 处。

注意：迷宫和房间的构造函数如下：

public Wall(int r1, int r2)
{
    room1 = r1;
    room2 = r2;
}

public Room(int i)
{
    id = i;
    distance = -1;
    paths = new Vector<Integer>();
}

如果有人愿意建议一个可以在 Java 中运行的实现，我将不胜感激，谢谢。

Answer 1

首先，我真的很喜欢迷宫的想法，并且一直在用 Java 开发一个类似的项目来生成Torus Mazes。

要生成迷宫，您需要查看以下关键语句：

...每个房间都有一个名为路径的邻接列表，并且 Room 类有一个变量 id，用于标识每个图的顶点

这告诉我们什么？它告诉我们我们需要的数据结构！当您处理此类问题时；由边连接的相邻向量，您无疑将创建一个邻接列表。

有几种不同的方法可以做到这一点，但到目前为止，最简单的（并且在这种情况下可以说是最有效的）是创建一个链表数组。这是我在没有使用 Java 库中的内置结构的情况下创建的邻接列表，但如果您选择使用内置结构，则可以使用相同的逻辑LinkedList<>。

/*
 * The Node class creates individual elements that populate the 
 * List class. Contains indexes of the node's neighbors and their
 * respective edge weights
 */
class Node {
    public int top;
    public int topWeight;
    public int bottom;
    public int bottomWeight;
    public int left;
    public int leftWeight;
    public int right;
    public int rightWeight;
    public int numConnec;

    // Default constructor, ititializes neghbors to -1 by default and edge
    // weights to 0
    Node () {
        top    = -1;
        right  = -1;
        bottom = -1;
        left   = -1;
    }
} // End Node class


/*
 * The List class contains Nodes, which are linked to one another
 * to create a Linked List. Used as an adjacency list in the
 * UnionFind class
 */
 class List {
    public Node neighbors;

    // Default constructor
    List () {
        neighbors = new Node ();
    }

    /**
     * Generates valid edges for the node, also assigns a randomly generated weight to that edge
     * @param i         The row that the node exists on, used to generate outer-node edges
     * @param j         The index of the node in the maze from 0 to (2^p)^2 - 1
     * @param twoP      Represents the dimensions of the maze, used in calculating valid edges
     * @param choice    Randomly generated number to choose which edge to generate
     * @param weight    Randomly generated number to assign generated edge a weight
     * @return          If the assignment was done correctly, returns true. Else returns false.
     */
    public boolean validEdges (int i, int j, int twoP, int choice, int weight) {
        if (neighbors.numConnec < 4) {
            // Top
            if (choice == 0) {
                neighbors.top = generateTop(i, j, twoP);
                neighbors.topWeight = weight;
                neighbors.numConnec++;
            }

            // Right
            else if (choice == 1) {
                neighbors.right = generateRight(i, j, twoP);
                neighbors.rightWeight = weight;
                neighbors.numConnec++;
            }

            // Bottom
            else if (choice == 2) {
                neighbors.bottom = generateBottom(i, j, twoP);
                neighbors.bottomWeight = weight;
                neighbors.numConnec++;
            }

            // Left
            else if (choice == 3) {
                neighbors.left = generateLeft(i, j, twoP);
                neighbors.leftWeight = weight;
                neighbors.numConnec++;
            }
        }
        else {
            return false;
        }
        return true;
    }

    public int generateTop (int i, int j, int twoP) {
        int neighbor = 0;

        // Set the top neighbor
        if (i == 0) {
            neighbor = j + twoP * (twoP + (-1));
        }
        else {
            neighbor = j + (-twoP);
        }
        return neighbor;
    }

    public int generateRight (int i, int j, int twoP) {
        int neighbor = 0;

        // Set the right neighbor
        if (j == twoP * (i + 1) + (-1)) {
            neighbor = twoP * i;
        }
        else {
            neighbor = j + 1;
        }
        return neighbor;
    }

    public int generateBottom (int i, int j, int twoP) {
        int neighbor = 0;

        // Set the bottom neighbor
        if (i == twoP + (-1)) {
            neighbor = j - twoP * (twoP + (-1));
        }
        else {
            neighbor = j + twoP;
        }
        return neighbor;
    }

    public int generateLeft (int i, int j, int twoP) {
        int neighbor = 0;

        // Set the left neighbor
        if (j == twoP * i) {
            neighbor = twoP * (i + 1) + (-1);
        }
        else {
            neighbor = j + (-1);
        }
        return neighbor;
    }
} // End List class

为了解决这个迷宫，这听起来像是一个 Dijkstra 算法的实现可以解决的问题。

Dijkstra 的工作方式是从您的第一个节点开始创建已知集合。然后，您确定到下一条边的最短路径并将该节点添加到已知集合中。每次寻找下一条最短路径时，都会添加从第一个节点经过的距离。

这个过程一直持续到所有节点都在已知集合中并且已经计算出到达目标的最短路径。

最短路径