Question

我正在尝试使用这种算法来检测网格上的相似组件。这是一个简单的游戏演示，可以在12x10网格上随机丢弃碎片，然后在每个碎片检查网格后检查三个或更多相邻碎片的任何组。我正在使用以下代码尝试在没有泛洪填充，递归或堆栈/队列的情况下执行此操作。它似乎几乎可以工作，但有时会破坏不同类型的方块，或者留下应该被摧毁的方块。那么，算法的逻辑是错误的，还是实现/编码错了？

编辑：我认为现在有效。见评论

public void checkMatches(int type)
{
    /*
     * Step 1: Iterate through each square to see how many of the same type are adjacent to it
     */
    for (int i = 0; i < PIECES_WIDE; i++)
    {
        for (int j = 0; j < PIECES_TALL; j++)
        {
            if (grid[i][j].getType() == type) // EDITED IN CODE. Make sure current square is of correct type
            {
                if (i > 0) // Bounds checking
                    if (grid[i - 1][j].getType() == type)
                        grid[i][j].setAdj(grid[i][j].getAdj() + 1);
                if (i < PIECES_WIDE - 1) // Bounds checking
                    if (grid[i + 1][j].getType() == type)
                        grid[i][j].setAdj(grid[i][j].getAdj() + 1);
                if (j > 0) // Bounds checking
                    if (grid[i][j - 1].getType() == type)
                        grid[i][j].setAdj(grid[i][j].getAdj() + 1);
                if (j < PIECES_TALL - 1) // Bounds checking
                    if (grid[i][j + 1].getType() == type)
                        grid[i][j].setAdj(grid[i][j].getAdj() + 1);
            }
        }
    }

    /*
     * Step 2: If there are 2 or more adjacent squares with the same type then it is part of a blob and to be destroyed
     */
    for (int i = 0; i < PIECES_WIDE; i++)
    {
        for (int j = 0; j < PIECES_TALL; j++)
        {
            if (grid[i][j].getAdj() >= 2)
                grid[i][j].setDestroy(true);
        }
    }

    /*
     * Step 3: If there is only 1 adjacent, then check to see if any adjacent squares have been marked to be destroyed (part
     * of a group). If so, set these to be destroyed as well.
     */
    for (int i = 0; i < PIECES_WIDE; i++)
    {
        for (int j = 0; j < PIECES_TALL; j++)
        {
            if (grid[i][j].getAdj() == 1)
            {
                if (i > 0) // Bounds checking
                    if (grid[i - 1][j].isDestroy() == true)
                    {
                        grid[i][j].setDestroy(true);
                        break;
                    }
                if (i < PIECES_WIDE - 1) // Bounds checking
                    if (grid[i + 1][j].isDestroy() == true)
                    {
                        grid[i][j].setDestroy(true);
                        break;
                    }
                if (j > 0) // Bounds checking
                    if (grid[i][j - 1].isDestroy() == true)
                    {
                        grid[i][j].setDestroy(true);
                        break;
                    }
                if (j < PIECES_TALL - 1) // Bounds checking
                    if (grid[i][j + 1].isDestroy() == true)
                    {
                        grid[i][j].setDestroy(true);
                        break;
                    }
            }
        }
    }

    /*
     * Step 4: Iterate through grid and destroy the squares marked for destruction and reset all squares to 0 adjacent and
     * destroy flag to false
     */
    for (int i = 0; i < PIECES_WIDE; i++)
    {
        for (int j = 0; j < PIECES_TALL; j++)
        {
            if (grid[i][j].isDestroy())
                destroyPiece(grid[i][j]);

            grid[i][j].setAdj(0);
            grid[i][j].setDestroy(false);
        }
    }
}

Answer 1

将项目排序到存储桶时，可以更快地找到组。您可以使用空间索引，例如kd-tree。

检测没有递归或堆栈/队列的组

1 个答案: