这实际上是我们的论文中,我们需要使用道格拉斯 - 普克算法在简化的线条,可以anyboy帮助我如何实现这一个Android应用程序。Android:如何从绘制的线中获取点的字符串?
我只是想知道如何让我画从线点串并通过减少总没有简化线。基于下面给出的代码点?
这是主类。
public class SketchTimeNewActivity extends GraphicsView implements ColorOption.OnColorChangedListener {
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(new MyView(this));
myPaint = new Paint();
myPaint.setAntiAlias(true);
myPaint.setDither(true);
myPaint.setColor(Color.CYAN);
myPaint.setStyle(Paint.Style.STROKE);
myPaint.setStrokeJoin(Paint.Join.ROUND);
myPaint.setStrokeCap(Paint.Cap.ROUND);
myPaint.setStrokeWidth(12);
}
private Paint myPaint;
public void colorChanged(int color) {
myPaint.setColor(color);
}
public class MyView extends View {
private static final float MINP = 0.25f;
private static final float MAXP = 0.75f;
private Bitmap mBitmap;
private Canvas mCanvas;
private Path mPath;
private Paint mBitmapPaint;
public MyView(Context c) {
super(c);
mPath = new Path();
mBitmapPaint = new Paint(Paint.DITHER_FLAG);
}
@Override
protected void onSizeChanged(int width, int height, int oldwidth, int oldheight) {
super.onSizeChanged(width, height, oldwidth, oldheight);
mBitmap = Bitmap.createBitmap(width, height, Bitmap.Config.ARGB_8888);
mCanvas = new Canvas(mBitmap);
}
@Override
protected void onDraw(Canvas canvas) {
canvas.drawColor(color.black);
canvas.drawBitmap(mBitmap, 0, 0, mBitmapPaint);
canvas.drawPath(mPath, myPaint);
}
private float mX, mY;
private static final float TOUCH_TOLERANCE = 4;
private void touch_start(float x, float y) {
mPath.reset();
mPath.moveTo(x, y);
mX = x;
mY = y;
}
private void touch_move(float x, float y) {
float dx = Math.abs(x - mX);
float dy = Math.abs(y - mY);
if (dx >= TOUCH_TOLERANCE || dy >= TOUCH_TOLERANCE) {
mPath.quadTo(mX, mY, (x + mX)/2, (y + mY)/2);
mX = x;
mY = y;
}
}
private void touch_up() {
mPath.lineTo(mX, mY);
// commit the path to our offscreen
mCanvas.drawPath(mPath, myPaint);
// kill this so we don't double draw
mPath.reset();
}
@Override
public boolean onTouchEvent(MotionEvent event) {
float x = event.getX();
float y = event.getY();
switch (event.getAction()) {
case MotionEvent.ACTION_DOWN:
touch_start(x, y);
invalidate();
break;
case MotionEvent.ACTION_MOVE:
touch_move(x, y);
invalidate();
break;
case MotionEvent.ACTION_UP:
touch_up();
invalidate();
break;
}
return true;
}
}
private static final int COLOR_MENU_ID = Menu.FIRST;
private static final int EXISTING_MENU_ID = Menu.FIRST + 2;
private static final int ENHANCED_MENU_ID = Menu.FIRST + 3;
private static final int ERASE_MENU_ID = Menu.FIRST + 1;
@Override
public boolean onCreateOptionsMenu(Menu menu) {
super.onCreateOptionsMenu(menu);
menu.add(0, COLOR_MENU_ID, 0, "Color").setShortcut('1', 'c');
menu.add(0, EXISTING_MENU_ID, 0, "Enhanced").setShortcut('2', 's');
menu.add(0, ENHANCED_MENU_ID, 0, "Existing").setShortcut('3', 'z');
menu.add(0, ERASE_MENU_ID, 0, "Erase").setShortcut('4', 'z');
return true;
}
@Override
public boolean onPrepareOptionsMenu(Menu menu) {
super.onPrepareOptionsMenu(menu);
return true;
}
@Override
public boolean onOptionsItemSelected(MenuItem item) {
myPaint.setXfermode(null);
myPaint.setAlpha(0xFFAAAAAA);
当点击了现有的菜单,这将简化被绘制的线,并显示具有更小的点,或者已经简化一个线的线。我打算为它创建一个新类,但我不知道如何从画布中绘制的线中获取点串。
switch (item.getItemId()) {
case COLOR_MENU_ID:
new ColorOption(this, this, myPaint.getColor()).show();
return true;
/** case EXISTING_MENU_ID:
return true;
case ENHANCED_MENU_ID:
return true;*/
case ERASE_MENU_ID:{
myPaint.setColor(color.black);
myPaint.setXfermode(new PorterDuffXfermode(PorterDuff.Mode.CLEAR));
return true;
}
}
return super.onOptionsItemSelected(item);
}
}
您正在从'onTouchEvent'生成线上的点,所以不是试图在之后查询Canvas,为什么不简单地保留这些创建点的列表?您可以在绘制每个新线段时添加一个点。 – 2011-12-23 07:52:06
非常感谢您的帮助。你能否请进一步解释我将如何去做?非常感谢, – user1081908 2011-12-23 14:29:44
当你说这是你的论文时,你的意思是说它是作业吗? – Zoot 2011-12-23 16:29:20