#include "opencv2/opencv.hpp"
#include
#include
#include

using namespace cv;
using namespace std;

int thresh = 50, N = 11;
const char* wndname = "Square Detection Demo";

int calcdistance(Point_ &point1, Point_ &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;  

}

// 查找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=(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 >& squares )
{
int h = image.rows;
int w = image.cols;
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, Size(image.cols/2, image.rows/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找正方形  
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 >& 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);
// 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> squares;

findSquares(image, squares);

for (auto itx = squares.begin(); itx != squares.end(); 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();  
return 0;  

}