3

我无法有效地实现 Eric Lippert 推荐的这种通用方法。他的博客概述了一种非常简单有效的创建 A Star 算法的方法（可在此处找到）。这是快速运行。

实际路径查找的代码：

class Path<Node> : IEnumerable<Node>
{
    public Node LastStep { get; private set; }
    public Path<Node> PreviousSteps { get; private set; }
    public double TotalCost { get; private set; }
    private Path(Node lastStep, Path<Node> previousSteps, double totalCost)
    {
        LastStep = lastStep;
        PreviousSteps = previousSteps;
        TotalCost = totalCost;
    }
    public Path(Node start) : this(start, null, 0) { }
    public Path<Node> AddStep(Node step, double stepCost)
    {
        return new Path<Node>(step, this, TotalCost + stepCost);
    }
    public IEnumerator<Node> GetEnumerator()
    {
        for (Path<Node> p = this; p != null; p = p.PreviousSteps)
            yield return p.LastStep;
    }
    IEnumerator IEnumerable.GetEnumerator()
    {
        return this.GetEnumerator();
    }
}

class AStar
{

    static public Path<Node> FindPath<Node>(
        Node start,
        Node destination,
        Func<Node, Node, double> distance,
        Func<Node, double> estimate)
        where Node : IHasNeighbours<Node>
    {
        var closed = new HashSet<Node>();
        var queue = new PriorityQueue<double, Path<Node>>();
        queue.Enqueue(0, new Path<Node>(start));
        while (!queue.IsEmpty)
        {
            var path = queue.Dequeue();
            if (closed.Contains(path.LastStep))
                continue;
            if (path.LastStep.Equals(destination))
                return path;
            closed.Add(path.LastStep);
            foreach (Node n in path.LastStep.Neighbours)
            {
                double d = distance(path.LastStep, n);
                if (n.Equals(destination))
                    d = 0;
                var newPath = path.AddStep(n, d);
                queue.Enqueue(newPath.TotalCost + estimate(n), newPath);
            }
        }
        return null;
    }

    /// <summary>
    /// Finds the distance between two points on a 2D surface.
    /// </summary>
    /// <param name="x1">The IntPoint on the x-axis of the first IntPoint</param>
    /// <param name="x2">The IntPoint on the x-axis of the second IntPoint</param>
    /// <param name="y1">The IntPoint on the y-axis of the first IntPoint</param>
    /// <param name="y2">The IntPoint on the y-axis of the second IntPoint</param>
    /// <returns></returns>
    public static long Distance2D(long x1, long y1, long x2, long y2)
    {
        //     ______________________
        //d = &#8730; (x2-x1)^2 + (y2-y1)^2
        //

        //Our end result
        long result = 0;
        //Take x2-x1, then square it
        double part1 = Math.Pow((x2 - x1), 2);
        //Take y2-y1, then sqaure it
        double part2 = Math.Pow((y2 - y1), 2);
        //Add both of the parts together
        double underRadical = part1 + part2;
        //Get the square root of the parts
        result = (long)Math.Sqrt(underRadical);
        //Return our result
        return result;
    }

    /// <summary>
    /// Finds the distance between two points on a 2D surface.
    /// </summary>
    /// <param name="x1">The IntPoint on the x-axis of the first IntPoint</param>
    /// <param name="x2">The IntPoint on the x-axis of the second IntPoint</param>
    /// <param name="y1">The IntPoint on the y-axis of the first IntPoint</param>
    /// <param name="y2">The IntPoint on the y-axis of the second IntPoint</param>
    /// <returns></returns>
    public static int Distance2D(int x1, int y1, int x2, int y2)
    {
        //     ______________________
        //d = &#8730; (x2-x1)^2 + (y2-y1)^2
        //

        //Our end result
        int result = 0;
        //Take x2-x1, then square it
        double part1 = Math.Pow((x2 - x1), 2);
        //Take y2-y1, then sqaure it
        double part2 = Math.Pow((y2 - y1), 2);
        //Add both of the parts together
        double underRadical = part1 + part2;
        //Get the square root of the parts
        result = (int)Math.Sqrt(underRadical);
        //Return our result
        return result;
    }

    public static long Distance2D(Point one, Point two)
    {
        return AStar.Distance2D(one.X, one.Y, two.X, two.Y);
    }
}

优先队列代码：

class PriorityQueue<P, V>
{
    private SortedDictionary<P, Queue<V>> list = new SortedDictionary<P, Queue<V>>();
    public void Enqueue(P priority, V value)
    {
        Queue<V> q;
        if (!list.TryGetValue(priority, out q))
        {
            q = new Queue<V>();
            list.Add(priority, q);
        }
        q.Enqueue(value);
    }
    public V Dequeue()
    {
        // will throw if there isn’t any first element!
        var pair = list.First();
        var v = pair.Value.Dequeue();
        if (pair.Value.Count == 0) // nothing left of the top priority.
            list.Remove(pair.Key);
        return v;
    }
    public bool IsEmpty
    {
        get { return !list.Any(); }
    }
}

以及获取附近节点的接口：

interface IHasNeighbours<N>
{
    IEnumerable<N> Neighbours { get; }
}

这是我无法有效实施的部分。我可以创建一个能够被路径查找使用的类，但是查找附近的节点变得很痛苦。本质上，我最终要做的是创建一个类，在这种情况下，它被视为单个图块。但是，为了获取所有附近的节点，我必须将一个值传递给该图块，其中包含所有其他图块的列表。这非常麻烦，让我相信一定有更简单的方法。

这是我使用 System.Drawing.Point 包装器的实现：

class TDGrid : IHasNeighbours<TDGrid>, IEquatable<TDGrid>
{
    public Point GridPoint;
    public List<Point> _InvalidPoints = new List<Point>();
    public Size _GridSize = new Size();
    public int _GridTileSize = 50;

    public TDGrid(Point p, List<Point> invalidPoints, Size gridSize)
    {
        GridPoint = p;
        _InvalidPoints = invalidPoints;
        _GridSize = gridSize;
    }

    public TDGrid Up(int gridSize)
    {
        return new TDGrid(new Point(GridPoint.X, GridPoint.Y - gridSize));
    }
    public TDGrid Down(int gridSize)
    {
        return new TDGrid(new Point(GridPoint.X, GridPoint.Y + gridSize));
    }
    public TDGrid Left(int gridSize)
    {
        return new TDGrid(new Point(GridPoint.X - gridSize, GridPoint.Y));
    }
    public TDGrid Right(int gridSize)
    {
        return new TDGrid(new Point(GridPoint.X + gridSize, GridPoint.Y));
    }

    public IEnumerable<TDGrid> IHasNeighbours<TDGrid>.Neighbours
    {
        get { return GetNeighbours(this); }
    }

    private List<TDGrid> GetNeighbours(TDGrid gridPoint)
    {
        List<TDGrid> retList = new List<TDGrid>();
        if (IsGridSpotAvailable(gridPoint.Up(_GridTileSize)))
            retList.Add(gridPoint.Up(_GridTileSize)); ;
        if (IsGridSpotAvailable(gridPoint.Down(_GridTileSize)))
            retList.Add(gridPoint.Down(_GridTileSize));
        if (IsGridSpotAvailable(gridPoint.Left(_GridTileSize)))
            retList.Add(gridPoint.Left(_GridTileSize));
        if (IsGridSpotAvailable(gridPoint.Right(_GridTileSize)))
            retList.Add(gridPoint.Right(_GridTileSize));
        return retList;
    }

    public bool IsGridSpotAvailable(TDGrid gridPoint)
    {
        if (_InvalidPoints.Contains(gridPoint.GridPoint))
            return false;

        if (gridPoint.GridPoint.X < 0 || gridPoint.GridPoint.X > _GridSize.Width)
            return false;
        if (gridPoint.GridPoint.Y < 0 || gridPoint.GridPoint.Y > _GridSize.Height)
            return false;

        return true;
    }

    public override int GetHashCode()
    {
        return GridPoint.GetHashCode();
    }

    public override bool Equals(object obj)
    {
        return this.GridPoint == (obj as TDGrid).GridPoint;
    }

    public bool Equals(TDGrid other)
    {
        return this.GridPoint == other.GridPoint;
    }
}

List _InvalidPoints 是我失败的地方。我可以将它传递给每个创建的 TDGrid，但考虑到所有其余代码的简单程度，这似乎是一种巨大的资源浪费。我知道这是我缺乏知识，但我无法搜索它。

必须有另一种实现方式：

interface IHasNeighbours<N>
{
    IEnumerable<N> Neighbours { get; }
}

有人对此有任何想法吗？

编辑——这是路径查找代码：

    public void FindPath(TDGrid start, TDGrid end)
    {
        AStar.FindPath<TDGrid>(start, end, (p1, p2) => { return AStar.Distance2D(p1.GridPoint, p2.GridPoint); }, (p1) => { return AStar.Distance2D(p1.GridPoint, end.GridPoint); });
    }
4

2 回答 2

1

听起来您在这里有两个不同的问题。您需要表示节点和节点本身之间的路径。您可能会发现分别表示这两个概念最容易。

例如，在下面的代码中，Grid 类跟踪节点的连接方式。这可以像存储作为墙壁（即被阻挡）的瓷砖的散列集一样简单。要确定一个节点是否可达，请检查它是否在哈希集中。这只是一个简单的例子。还有许多其他方式可以表示图表，请参阅Wikipedia。

单个节点可以表示为 Grid 上的两个坐标，只需要三个值：行、列和 Grid 本身。这允许动态创建每个单独的节点（享元模式）。

希望有帮助！

class Grid
{
    readonly int _rowCount;
    readonly int _columnCount;

    // Insert data for master list of obstructed cells
    // or master list of unobstructed cells

    public Node GetNode(int row, int column)
    {
        if (IsOnGrid(row, column) && !IsObstructed(row, column))
        {
            return new Node(this, row, column);
        }

        return null;
    }

    private bool IsOnGrid(int row, int column)
    {
        return row >= 0 && row < _rowCount && column >= 0 && column < _columnCount;
    }

    private bool IsObstructed(int row, int column)
    {
        // Insert code to check whether specified row and column is obstructed
    }
}

class Node : IHasNeighbours<Node>
{
    readonly Grid _grid;
    readonly int _row;
    readonly int _column;

    public Node(Grid grid, int row, int column)
    {
        _grid = grid;
        _row = row;
        _column = column;
    }

    public Node Up
    {
        get
        {
            return _grid.GetNode(_row - 1, _column);
        }
    }

    public Node Down
    {
        get
        {
            return _grid.GetNode(_row + 1,_column);
        }
    }

    public Node Left
    {
        get
        {
            return _grid.GetNode(_row, _column - 1);
        }
    }

    public Node Right
    {
        get
        {
            return _grid.GetNode(_row, _column + 1);
        }
    }

    public IEnumerable<Node> Neighbours
    {
        get
        {
            Node[] neighbors = new Node[] {Up, Down, Left, Right};
            foreach (Node neighbor in neighbors)
            {
                if (neighbor != null)
                {
                    yield return neighbor;
                }
            }
        }
    }
}
于 2012-06-11T21:53:13.130 回答
0

这是我最终使用的实现，与 Special Touch 的解决方案非常相似。SpacialObject 是一个点。

public class Tile : SpacialObject, IHasNeighbours<Tile>
{
    public Tile(int x, int y)
        : base(x, y)
    {
        CanPass = true;
    }

    public bool CanPass { get; set; }

    public Point GetLocation(int gridSize)
    {
        return new Point(this.X * gridSize, this.Y * gridSize);
    }

    public IEnumerable<Tile> AllNeighbours { get; set; }
    public IEnumerable<Tile> Neighbours { get { return AllNeighbours.Where(o => o.CanPass); } }

    public void FindNeighbours(Tile[,] gameBoard)
    {
        var neighbours = new List<Tile>();

        var possibleExits = X % 2 == 0 ? EvenNeighbours : OddNeighbours;
        possibleExits = GetNeighbours;

        foreach (var vector in possibleExits)
        {
            var neighbourX = X + vector.X;
            var neighbourY = Y + vector.Y;

            if (neighbourX >= 0 && neighbourX < gameBoard.GetLength(0) && neighbourY >= 0 && neighbourY < gameBoard.GetLength(1))
                neighbours.Add(gameBoard[neighbourX, neighbourY]);
        }

        AllNeighbours = neighbours;
    }

    public static List<Point> GetNeighbours
    {
        get
        {
            return new List<Point>
            {
                new Point(0, 1),
                new Point(1, 0),
                new Point(0, -1),
                new Point(-1, 0),
            };
        }
    }

    public static List<Point> EvenNeighbours
    {
        get
        {
            return new List<Point>
            {
                new Point(0, 1),
                new Point(1, 1),
                new Point(1, 0),
                new Point(0, -1),
                new Point(-1, 0),
                new Point(-1, 1),
            };
        }
    }

    public static List<Point> OddNeighbours
    {
        get
        {
            return new List<Point>
            {
                new Point(0, 1),
                new Point(1, 0),
                new Point(1, -1),
                new Point(0, -1),
                new Point(-1, 0),
                new Point(-1, -1),
            };
        }
    }
}

然后在我使用的主程序中：

    private void InitialiseGameBoard()
    {
        GameBoard = new Tile[_Width, _Height];

        for (var x = 0; x < _Width; x++)
        {
            for (var y = 0; y < _Height; y++)
            {
                GameBoard[x, y] = new Tile(x, y);
            }
        }

        AllTiles.ToList().ForEach(o => o.FindNeighbours(GameBoard));

        int startX = 0, endX = GameBoard.GetLength(0) - 1;
        int startEndY = GameBoard.GetLength(1) / 2;
        _StartGridPoint = new Point(startX, startEndY);
        _EndGridPoint = new Point(endX, startEndY);
        //GameBoard[startX, startEndY].CanPass = false;
        //GameBoard[endX, startEndY].CanPass = false;
    }

    private void BlockOutTiles()
    {
        GameBoard[2, 5].CanPass = false;
        GameBoard[2, 4].CanPass = false;
        GameBoard[2, 2].CanPass = false;
        GameBoard[3, 2].CanPass = false;
        GameBoard[4, 5].CanPass = false;
        GameBoard[5, 5].CanPass = false;
        GameBoard[5, 3].CanPass = false;
        GameBoard[5, 2].CanPass = false;
    }

    public IEnumerable<Tile> AllTiles
    {
        get
        {
            for (var x = 0; x < _Width; x++)
                for (var y = 0; y < _Height; y++)
                    yield return GameBoard[x, y];
        }
    }
于 2012-06-16T16:36:51.983 回答