扫雷清除算法

时间:2014-04-19 03:18:33

标签: algorithm actionscript-3 minesweeper

我正在尝试在闪光灯中为一个项目重新创建扫雷,我已经获得了放置地雷和数字的权利。我似乎无法弄清楚用于扩展点击图块和清除其邻居的递归算法。有两个数组,_map包含背景+地雷和数字,_clickable map保存位于背景顶部的图块。所以基本上,我要求的是递归的帮助。如果这还不够清楚,我会用必要的信息更新问题。

private function onClick(e:MouseEvent):void 
{
    trace("on Click");

    var symbol = e.currentTarget;
    var tempTileMask:TileMask = symbol;
    var tempTile:Tile = _map[tempTileMask.yCoord][tempTileMask.xCoord]

    if (tempTile.hasMine)
    {
        for (var i:int = _gridSize - 1; i >= 0; i--)
            for (var j:int = _gridSize - 1; j >= 0; j--)
            {
                var temp:TileMask = _clickableMap[i][j];
                removeChild(temp);
            }

        //explosion();

        //gameOver();
    }

    if (tempTile.hasNumber)
    {
        removeChild(tempTileMask);
        tempTile.cleared = true;
    }

    else
    {
        clearTiles(tempTile.xCoord, tempTile.yCoord);
    }
}

这是我修改以清除地雷的功能

function clearTiles( x:int, y:int )
{
    // Get an object that contains the tile information for the location
    // we are checking
    if(_map[y][x] != null) 
    {
        var tempTile:Tile = _map[y][x];
        var tempTileMask:TileMask = _clickableMap[y][x];

        // Check if the location we are checking is out of the bounds of the
        // playable area

        trace(tempTile);
        if ( tempTile.outOfBounds(x, y) )
        {
            return;
        }

        // If the tile has already been revealed then there is nothing to do

        trace("before clearing");
        if ( tempTile.cleared )
        {
            trace("clearing tile");
            return;
        }

        // If the tile hasn't been revealed and it's an empty square we
        // reveal the location then call this function again for each
        // surrounding block

        trace("before check for surrounding tiles");
        if ( tempTile.hasNumber != true && tempTile.hasMine != true )
        {
            trace("check for surrounding tiles");

            // Remove the mask hiding the tiles property
            removeChild( tempTileMask );
            // Set the tile as cleared
            tempTile.cleared = true;

            if(_map[tempTile.yCoord][tempTile.xCoord - 1] != null)
            {
                var tile1:Tile =_map[tempTile.yCoord][tempTile.xCoord - 1]

                if(!tile1.cleared)
                    clearTiles( tempTile.xCoord - 1 , tempTile.yCoord ); //Check tile to the left
            }

            if(_map[tempTile.yCoord][tempTile.xCoord + 1] != null)
            {
                var tile2:Tile =_map[tempTile.yCoord][tempTile.xCoord + 1]

                if(!tile2.cleared)
                    clearTiles( tempTile.xCoord + 1 , tempTile.yCoord ); //Check tile to the left
            }

            if(_map[tempTile.yCoord - 1][tempTile.xCoord] != null)
            {
                var tile3:Tile =_map[tempTile.yCoord - 1][tempTile.xCoord]

                if(!tile3.cleared)
                    clearTiles( tempTile.xCoord, tempTile.yCoord - 1 ); //Check tile to the left
            }

            if(_map[tempTile.yCoord + 1][tempTile.xCoord] != null)
            {
                var tile4:Tile =_map[tempTile.yCoord + 1][tempTile.xCoord]

                if(!tile4.cleared)
                    clearTiles( tempTile.xCoord, tempTile.yCoord + 1 ); //Check tile to the left
            }

            if(_map[tempTile.yCoord - 1][tempTile.xCoord - 1] != null)
            {
                var tile5:Tile =_map[tempTile.yCoord - 1][tempTile.xCoord - 1]

                if(!tile5.cleared)
                    clearTiles( tempTile.xCoord - 1, tempTile.yCoord - 1 ); //Check tile to the left
            }

            if(_map[tempTile.yCoord + 1][tempTile.xCoord + 1] != null)
            {
                var tile6:Tile =_map[tempTile.yCoord + 1][tempTile.xCoord + 1]

                if(!tile6.cleared)
                    clearTiles( tempTile.xCoord + 1, tempTile.yCoord + 1 ); //Check tile to the left
            } 

            if(_map[tempTile.yCoord - 1][tempTile.xCoord + 1] != null)
            {
                var tile7:Tile =_map[tempTile.yCoord - 1][tempTile.xCoord + 1]

                if(!tile7.cleared)
                    clearTiles( tempTile.xCoord + 1, tempTile.yCoord - 1 ); //Check tile to the left
            }


            if(_map[tempTile.yCoord + 1][tempTile.xCoord - 1] != null)
            {
                var tile8:Tile =_map[tempTile.yCoord + 1][tempTile.xCoord - 1]

                if(!tile8.cleared)
                    clearTiles( tempTile.xCoord - 1, tempTile.yCoord + 1 ); //Check tile to the left
            }
        }
    }
    else
        return;

}

1 个答案:

答案 0 :(得分:4)

如果我搞砸了actionscript语法,请原谅。我只是喜欢递归函数,所以我不得不回答这个问题。我只是粘贴代码,其中包含我可以提出的最准确的评论。如果您有任何疑问,请询问。我确定您可以将所需的任何内容转换为相应的函数调用,属性引用等。

// Function to check if a tile is an empty space and then call
// itself for the surrounding tiles

function clearTiles( x:int, y:int ):void
{
    // Check if the location we are checking is out of the bounds of the
    // playable area
    if ( outOfBounds( x, y ) )
    {
        return;
    }

    // Get an object that contains the tile information for the location
    // we are checking
    var tempTile:Tile = _map[x][y];


    // If the tile has already been revealed then there is nothing to do
    if ( tempTile.cleared )
    {
        return;
    }

    // If the tile is a number then reveal it and return without checking
    // surrounding tiles
    var tempTileMask:DisplayObject = getTileMask(x,y);
    // since we're no longer in the click handler context, we need
    // to initialize the variable with something [TODO]
    if ( tempTile.hasNumber )
    {
        removeChild( tempTileMask );
        tempTile.cleared = true;
        return;
    }


    // If the tile hasn't been revealed and it's an empty square we
    // reveal the location then call this function again for each
    // surrounding block

    if ( tempTile.isEmpty )
    {
        // Remove the mask hiding the tiles property
        removeChild( tempTileMask );
        // Set the tile as cleared
        tempTile.cleared = true;

        clearTiles( tempTile.x - 1 , tempTile.y ); //Check tile to the left
        clearTiles( tempTile.x + 1 , tempTile.y ); //Check tile to the right
        clearTiles( tempTile.x , tempTile.y - 1 ); //Check tile above
        clearTiles( tempTile.x , tempTile.y + 1 ); //Check tile below
    }

}

您必须创建一个outOfBounds()函数来检查被检查的X和Y是否大于当前游戏板。看起来您将_gridSize引用为静态数字,因此我假设您的所有游戏板都是正方形(例如:4x4,9x9,120x120)。在这种情况下,您可以使用以下内容:

function outOfBounds( int x, int y )
{
    if ( x < 0 )
    {
        return true;
    }

    if ( y < 0 )
    {
        return true;
    }

    if ( x > _gridSize - 1 )
    {
        return true;
    }

    if ( y > _gridSize -1 )
    {
        return true;
    }

    return false;
}