[opencv]approxDP多边形逼近获取四边形轮廓信息#include "opencv2/opencv.hpp"
#include <iostream>
#include <math.h>
#include <string.h>
using namespace cv;
using namespace std;
int thresh = 50, N = 11;
const char* wndname = "Square Detection Demo";
int calcdistance(Point_<int> &point1, Point_<int> &point2) {
int x1 = point1.x;
int y1 = point1.y;
int x2 = point2.x;
int y2 = point2.y;
int dist = sqrt(pow(x2-x1,2)+pow(y2-x1,2));
return dist;
}
// 查向量之间的⾓度余弦
// 从pt0->pt1和从pt0->pt2
static double angle( Point pt1, Point pt2, Point pt0 )
{
double dx1 = pt1.x - pt0.x;
double dy1 = pt1.y - pt0.y;
double dx2 = pt2.x - pt0.x;
double dy2 = pt2.y - pt0.y;
return (dx1*dx2 + dy1*dy2)/sqrt((dx1*dx1 + dy1*dy1)*(dx2*dx2 + dy2*dy2) + 1e-10);
//cos<a,b>=(ab的内积)/(|a||b|)
}
// 返回图像上检测到的正⽅形序列
// 序列存储在指定的内存存储器中
static void findSquares( const Mat& image, vector<vector<Point> >& squares )
{
int h = ws;
int w = ls;
squares.clear();
Mat pyr, timg, gray0(image.size(), CV_8U), gray;
// 缩⼩和放⼤图像以滤除噪⾳
pyrDown(image, pyr, ls/2, ws/2));//⾼斯降噪，并只取奇数⾏列缩⼩图⽚
pyrUp(pyr, timg, image.size());//插⼊偶数⾏列，再次⾼斯降噪
vector<vector<Point> > contours;
// 在图像的每个颜⾊平⾯中查正⽅形
for( int c = 0; c < 3; c++ )
{
int ch[] = {c, 0};
mixChannels(&timg, 1, &gray0, 1, ch, 1);
/
/ 尝试⼏个阈值级别
for( int l = 0; l < N; l++ )
{
if( l == 0 )
{
Canny(gray0, gray, 0, thresh, 5);
dilate(gray, gray, Mat(), Point(-1,-1));
}
else
{
// apply threshold if l!=0:
/
/    tgray(x,y) = gray(x,y) < (l+1)*255/N ? 255 : 0
gray = gray0 >= (l+1)*255/N;
}
//注意_RETR_LIST参数指明最后只存了⾓点位置
findContours(gray, contours, RETR_LIST, CHAIN_APPROX_SIMPLE);
vector<Point> approx;
// test each contour
//⼀次取⼀个轮廓判断下是不是矩形
for( size_t i = 0; i < contours.size(); i++ )
{
//approx存储了近似后的轮廓
approxPolyDP(Mat(contours[i]), approx, arcLength(Mat(contours[i]), true)*0.02, true);
if( approx.size() == 4 && //矩形必须是四个点
fabs(contourArea(Mat(approx))) > 5000 &&
isContourConvex(Mat(approx)) )//必须是凸的，咋理解
{
double maxCosine = 0;
for( int j = 2; j < 5; j++ )
{
// find the maximum cosine of the angle between joint edges
double cosine = fabs(angle(approx[j%4], approx[j-2], approx[j-1]));
maxCosine = MAX(maxCosine, cosine);
}
//依次计算1,2,3顶点的⾓余弦，最⼤的余弦值，对应三个⾓中的最⼩⾓，也就是三个⾓中，最不像直⾓的                    //为什么是三个呢？三个中最不像直⾓的都接近直⾓了，剩下的⾃然也是直⾓
if( maxCosine < 0.3 )
if (calcdistance(approx[0],approx[3]) < h && calcdistance(approx[0],approx[1]) < w){
squares.push_back(approx);
}
}
}
}
}
}
// 函数绘制图像中的所有四边形
static void drawSquares(Mat& image, const vector<vector<Point> >& squares)
{
for( size_t i = 0; i < squares.size(); i++ )
{
const Point* p = &squares[i][0];
int n = (int)squares[i].size();
circle(image,squares[i][0],3,Scalar(0,0,255),1);
circle(image,squares[i][1],3,Scalar(0,0,255),1);
circle(image,squares[i][2],3,Scalar(0,0,255),1);
circle(image,squares[i][3],3,Scalar(0,0,255),1);
/rectangle函数opencv
/        polylines(image, &p, &n, 1, true, Scalar(0,255,0), 1, LINE_AA);
//        break;
Rect rr = boundingRect(squares[i]);
//通过最⼩包围盒来获取主要函数区域
vector<vector<Point>> main;
main[0][0] = squares[0][0];
main[0][1] = squares[0][1];
main[0][2] = squares[1][2];
main[0][3] = squares[1][3];
circle(image,main[0][0],1,Scalar(0,255,0),-1);
circle(image,main[0][1],1,Scalar(0,255,0),-1);
imshow("main",image);
RotatedRect minRect = minAreaRect(Mat(main[0]));
Point2f vertex[4];//⽤于存放最⼩矩形的四个顶点
minRect.points(vertex);//返回矩形的四个顶点给vertex
//绘制最⼩⾯积包围矩形
vector<Point>min_rectangle;
for (int i = 0; i < 4; i++)
{
line(image, vertex[i], vertex[(i + 1) % 4], Scalar(0, 255, 255), 1, 8);//⾮常巧妙的表达式
min_rectangle.push_back(vertex[i]);//将最⼩矩形的四个顶点以Point的形式放置在vector容器中        }
//        rectangle(image,Point(rr.x, rr.y), Point(rr.x + rr.width, rr.y + rr.height), 1);
/
/        float area = contourArea(squares[i], false);
//        cout << "area==" << area << endl;
//        break;
}
imshow(wndname, image);
}
int main(int/*argc*/, char** /*argv*/)
{
Mat image = imread("/home/leoxae/KeekoRobot/TestPic/⼤班/1.png");
vector<vector<Point>> squares;
findSquares(image, squares);
for (auto itx = squares.begin(); itx != d(); itx++){
vector<Point> points = *itx;
cout << "Pts=" << points << endl;
}
/*    Point b_tl = squares[0][0];
Point b_tr = squares[0][1];
Point b_bl = squares[0][2];
Point b_br = squares[0][3];
Point s_tl = squares[1][0];
Point s_tr = squares[1][1];
Point s_bl = squares[1][2];
Point s_br = squares[1][3];
cout << "b_tl==" << b_tl << endl;
cout << "b_tr==" << b_tr << endl;
cout << "b_bl==" << b_bl << endl;
cout << "b_br==" << b_br << endl;
cout << "s_tl==" << s_tl << endl;
cout << "s_tr==" << s_tr << endl;
cout << "s_bl==" << s_bl << endl;
cout << "s_br==" << s_br << endl;
circle(image,b_bl,3,Scalar(0,0,255),-1);
circle(image,b_tr,3,Scalar(0,0,255),-1);
circle(image,b_bl,3,Scalar(0,0,255),-1);
circle(image,b_br,3,Scalar(0,0,255),-1);
circle(image,s_tl,3,Scalar(0,255,0),-1);
circle(image,s_tr,3,Scalar(0,255,0),-1);
circle(image,s_bl,3,Scalar(0,255,0),-1);
circle(image,s_br,3,Scalar(0,255,0),-1);*/
//    imshow("drawcircle",image);
drawSquares(image, squares);
waitKey();
return0;
}

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