我有一个图像，我在其上使用了cvFindContours（） - 我将结果存储在imageContours中。然后我试图找到两个最大的轮廓并仅显示它们。但由于某种原因，它的显示方式超过2个轮廓。我的算法出了什么问题？感谢
编辑：哦，而且，在控制台中，我一直收到“错误”消息，如果没有任何选项是真的，我在下面编程。谁知道为什么？
cvFindContours (binMask, imageContoursMem, &imageContours, sizeof (CvContour), CV_RETR_EXTERNAL, CV_CHAIN_APPROX_SIMPLE); //find the contours in binMask, and store results in ImageContours
//following conditions used to set f_handContour to the bigger contour for the first time, and s_handContour to the smaller contour for the first time
if ( (cvContourArea (imageContours, CV_WHOLE_SEQ)) > (cvContourArea((imageContours -> h_next), CV_WHOLE_SEQ)))
{ //begin first contour bigger condition
f_handContour = imageContours;
s_handContour = (imageContours -> h_next);
} //end second contour bigger condition
else if ( (cvContourArea (imageContours, CV_WHOLE_SEQ)) < (cvContourArea((imageContours -> h_next), CV_WHOLE_SEQ)))
{ //begin second contour bigger condition
f_handContour = (imageContours -> h_next);
s_handContour = imageContours;
} //begin second contour biggger condition
else if ( (cvContourArea (imageContours, CV_WHOLE_SEQ)) == (cvContourArea((imageContours -> h_next), CV_WHOLE_SEQ)))
{ //begin contours equal condition
f_handContour = imageContours; //if contours equal assignment of contours doesn't matter
s_handContour = (imageContours -> h_next);
} //end contours equal condition
else
{ //begin error condition
cout<<"Error";
} //end error condition
while ((imageContours -> h_next) != NULL)
{ //start of biggest/second biggest contour recognition loop
imageContours = (imageContours -> h_next);
if ( cvContourArea (f_handContour, CV_WHOLE_SEQ) < cvContourArea (imageContours, CV_WHOLE_SEQ))
{ //start of next contour bigger than f_handContour condition
s_handContour = f_handContour;
f_handContour = imageContours;
} //end start of next contour bigger than f_handContour condition
else if ( cvContourArea (f_handContour, CV_WHOLE_SEQ) > cvContourArea (imageContours, CV_WHOLE_SEQ))
{ //start of next contour smaller than f_handContour condition
if ( cvContourArea (s_handContour, CV_WHOLE_SEQ) < cvContourArea (imageContours, CV_WHOLE_SEQ))
{ //startof next contour bigger than s_handContour condition
s_handContour = imageContours;
} //end of next contour bigger than s_handContour condition
else if ( cvContourArea (s_handContour, CV_WHOLE_SEQ) > cvContourArea (imageContours, CV_WHOLE_SEQ))
{ //start of next contour smaller than s_handContour condition
//just pass on contour
} //end of next contour smaller than s_handContour condition
else if ( cvContourArea (s_handContour, CV_WHOLE_SEQ) == cvContourArea (imageContours, CV_WHOLE_SEQ))
{ //start of next contour equal to s_handContour condition
//just pass on contour
} // end of next contour equal to s_handContour condition
else
{ //start of error condition
cout<<"Error";
} //end of error condition
} //end of next contour smaller than f_handContour Condition
else if ( cvContourArea (f_handContour, CV_WHOLE_SEQ) == cvContourArea (imageContours, CV_WHOLE_SEQ))
{ //start of next contour equal to f_handContour condition
if ( cvContourArea (s_handContour, CV_WHOLE_SEQ) < cvContourArea (imageContours, CV_WHOLE_SEQ))
{ //startof next contour bigger than s_handContour condition
s_handContour = imageContours;
} //end of next contour bigger than s_handContour condition
else
{ //start of error condition
cout<<"Error";
} //end of error condition
} //end of next contour equal to f_handContour condition
else
{ //start of error condition
cout<<"Error";
} //end of error condition
} //end of biggest/second biggest contour recognition loop
cvDrawContours (output, f_handContour, cvScalar (0,255,0), cvScalar (0,255,255), 1, 3,8, cvPoint (0,0)); //draws the first hand contour derived from binMask on ouput
cvDrawContours (output, s_handContour, cvScalar (0,255,0), cvScalar (0,255,255), 1, 3,8, cvPoint (0,0)); //draws the second hand contour derived from binMask on ouput
答案 0 :(得分：1)
CV_RETR_EXTERNAL仅检索极端外轮廓。 对于您的算法，您应该使用CV_RETR_LIST。 尝试使用max_level作为cvDrawContours函数的值0，因为值1绘制当前轮廓以及与它在同一级别上的所有其他轮廓。 你也可以很好地优化你的条件（这有点混乱），并且还将区域存储在一些局部变量中，以免过多地调用cvContourArea。