在具有透明度的图像中创建边界框

Question

我想用一种特定颜色裁剪图像，就像透明蒙版一样，我想创建一个最小的框来包围图像。

我的意思的图片肯定会更好地解释它：

RGB(255,0,255) 是透明蒙版，绿色框代表所需的边界框。

我已经算出了边界框的顶线：

    int nAlloc = (128 * 128) * 4;
    unsigned char* buf = new unsigned char[nAlloc];
    GetBitmapBits(hBMP, nAlloc, buf);
    for(int i = 0; i < nAlloc; i += 4)
    {
        if(buf[i] == 255 && buf[i + 1] == 0 && buf[i + 2] == 255)
        {
            continue;
        }
        // If I hit a first non-transparent pixel, 
        // I can then calculate the row where is that pixel located.
        else if(set_pixel == false)
        {
            set_pixel = true;
            index = ceil((float)(i / 4.0 / 128.0));
        }
        ... // Converting non-transparent pixels to Black&White
    }

    //I'm then drawing the bitmap to window like so:
    TransparentBlt(hdc, 5, 305 - index, 128, 128, hDC, 0, 0, 128, 128, RGB(255, 0, 255));

我想，我也知道如何确定最后一行，但我不确定，也不知道如何找出边界框的边。

Answer 1

要找到图像的边界框，您只需逐行和逐列遍历图像，直到找到不透明的像素。 通过这样做，您可以获得边界框的最小值和最大值。

RECT BoundingBox = { 0,0,0,0 };
const int nAlloc = (128 * 128) * 4;
unsigned char* buf = new unsigned char[nAlloc];
GetBitmapBits(hBMP, nAlloc, buf);

bool found;

//search upper bound
found = false;
for (int row = 0; row<128 && !found; row++) //row
{
    for (int col = 0; col<128 && !found; col++) //column
    {
        int idx = (row * 128 + col) * 4;
        if (!(buf[idx] == 255 && buf[idx + 1] == 0 && buf[idx + 2] == 255)) //not transparent
        {
            BoundingBox.top = row;
            found = true;
        }
    }
}

//search lower bound
found = false;
for (int row = 127; row >= 0 && !found; row--) //row
{
    for (int col = 127; col >= 0 && !found; col--) //column
    {
        int idx = (row * 128 + col) * 4;
        if (!(buf[idx] == 255 && buf[idx + 1] == 0 && buf[idx + 2] == 255)) //not transparent           {
            BoundingBox.bottom = row;
            found = true;
        }
    }
}


//search left bound
found = false;
for (int col = 0; col<128 && !found; col++) //row
{
    for (int row = 0; row<128 && !found; row++) //column
    {
        int idx = (row * 128 + col) * 4;
        if (!(buf[idx] == 255 && buf[idx + 1] == 0 && buf[idx + 2] == 255)) //not transparent           {
            BoundingBox.left = col;
            found = true;
        }
    }
}


//search right bound
found = false;
for (int col = 127; col >= 0 && !found; col--) //row
{
    for (int row = 127; row >= 0 && !found; row--) //column
    {
        int idx = (row * 128 + col) * 4;
        if (!(buf[idx] == 255 && buf[idx + 1] == 0 && buf[idx + 2] == 255)) //not transparent           {
            BoundingBox.right = col;
            found = true;
        }
    }
}

//now the variable "BoundingBox" contains your BoundingBox

希望我的回答对你有所帮助。

Answer 2

我不知道是否有一种优雅的方法可以做到这一点，但您可以简单地扫描整个图像，同时记录非透明像素的最小和最大列。

const int rowcnt = 128;
const int colcnt = 128;
const int bytesperpix = 4;
int pos = 0;
int colmin = colcnt;
int colmax = 0;
int colpos = 0;
for(int r = 0; r < rowcnt; ++r)
{
    for(int c = 0; c < colcnt * bytesperpix; c += 4)
    {
        pos = r * colcnt * bytesperpix + c;
        if(buf[pos] == 255 && buf[pos + 1] == 0 && buf[pos + 2] == 255)
        {
            continue;
        }
        else
        {
            colpos = c / bytesperpix;
            if(colpos < colmin)
                colmin = colpos;
            else if(colpos > colmax)
                colmax = colpos;
        }
    }
}