Question

可能重复

朋友你好，

我创建绘画应用程序，我有问题。如果我绘制没有填充的矩形和/或其他类似绑定区域并更改背景颜色，则矩形填充区域也会更改意味着整个画布颜色将填充新的背景颜色。如何保持背景或填充未绑定的画布区域，这里是图像

这是初始图片

enter image description here

更改获得此结果的背景颜色后

enter image description here

但如何这样做

enter image description here

Answer 1

final Point p1 = new Point();
                p1.x=(int) x; //x co-ordinate where the user touches on the screen
                p1.y=(int) y; //y co-ordinate where the user touches on the screen  
 new TheTask(yourbitmap, p1, sourceColor,targetColor).execute();// use asyntask for efficiency

 class TheTask extends AsyncTask<Void, Integer, Void> {

    Bitmap bmp;
    Point pt;
    int replacementColor,targetColor;
    ProgressDialog pd;
 public TheTask(Bitmap bm,Point p, int sc, int tc)
 {
this.bmp=bm;
this.pt=p;
this.replacementColor=tc;
this.targetColor=sc;
pd= new ProgressDialog(context);
pd.setMessage("Filling....");
    }
    @Override
    protected void onPreExecute() {
            pd.show();

    }

    @Override
    protected void onProgressUpdate(Integer... values) {

    }

    @Override
    protected Void doInBackground(Void... params) {
        FloodFill f= new FloodFill();
        f.floodFill(bmp,pt,targetColor,replacementColor);
        return null;
    }

    @Override
    protected void onPostExecute(Void result) { 
pd.dismiss();
invalidate();
    }

最后使用FloodFill算法填充封闭区域

    public class FloodFill {
public void floodFill(Bitmap  image, Point node, int targetColor,
        int replacementColor) {
    int width = image.getWidth();
    int height = image.getHeight();
    int target = targetColor;
    int replacement = replacementColor;
    if (target != replacement) {
        Queue<Point> queue = new LinkedList<Point>();
        do {
            int x = node.x;
            int y = node.y;
            while (x > 0 && image.getPixel(x - 1, y) == target) {
                x--;
            }
            boolean spanUp = false;
            boolean spanDown = false;
            while (x < width && image.getPixel(x, y) == target) {
                image.setPixel(x, y, replacement);
                if (!spanUp && y > 0 && image.getPixel(x, y - 1) == target) {
                    queue.add(new Point(x, y - 1));
                    spanUp = true;
                } else if (spanUp && y > 0
                        && image.getPixel(x, y - 1) != target) {
                    spanUp = false;
                }
                if (!spanDown && y < height - 1
                        && image.getPixel(x, y + 1) == target) {
                    queue.add(new Point(x, y + 1));
                    spanDown = true;
                } else if (spanDown && y < height - 1
                        && image.getPixel(x, y + 1) != target) {
                    spanDown = false;
                }
                x++;
            }
        } while ((node = queue.poll()) != null);
    }
}
}

enter image description here

Answer 2

这里是代码（你必须在触摸事件上绑定形状，否则它会改变形状的颜色）：

public class ttt extends View {
MyShape myShape;
public ttt(Context context) {
    super(context);
    // TODO Auto-generated constructor stub
    myShape = new MyShape();
    Paint paint = new Paint();
    paint.setColor(Color.WHITE);
    myShape.setPaint(paint);
}

@Override
protected void onDraw(Canvas canvas) {
    // TODO Auto-generated method stub
    super.onDraw(canvas);
    myShape.onDraw(canvas);     
}

@Override
public boolean onTouchEvent(MotionEvent event) {
    // TODO Auto-generated method stub
    int x = (int) event.getX();
    int y = (int) event.getY();

    switch (event.getAction()) {
    case MotionEvent.ACTION_DOWN:
        Paint paint = new Paint();
        paint.setColor(Color.BLUE);
        myShape.setPaint(paint);
        invalidate();
        break;

    default:
        break;
    }

    return super.onTouchEvent(event);
}

class MyShape {
    private Paint paint;
    public MyShape() {
        // TODO Auto-generated constructor stub
    }
    public void onDraw(Canvas canvas){
        canvas.drawCircle(15, 15, 30, getPaint());
    }
    /**
     * @param paint the paint to set
     */
    public void setPaint(Paint paint) {
        this.paint = paint;
    }
    /**
     * @return the paint
     */
    public Paint getPaint() {
        return paint;
    }

}

}

Answer 3

您可能还希望在以下两个问题中使用剪辑区域：

根据您的想法，它们可能很有用。

我找到了你的问题，因为我不知道谷歌的用途，我想要的答案是没有使用洪水填充。

填充整个画布，但保持绑定的填充区域，如圆形，矩形

3 个答案: