貝塞爾曲線詳解

貝塞爾曲線原理：
https://www.cnblogs.com/hnfxs/p/3148483.html
貝塞爾曲線的理解： 實際上就是利用了各個線段上等比例
https://www.jianshu.com/p/0c9b4b681724

貝塞爾曲線編程實現：
https://blog.csdn.net/Dang_boy/article/details/84570221

#include <iostream>
#include <opencv2/opencv.hpp>
#include <vector>
using namespace cv;
using std::cout;
using std::endl;
using std::vector;
vector<Point2f> bezierCurve(vector<Point2f> src);
int main(int argc, char const* argv[])
{
   while (1) {
       vector<Point2f> path;
       CvRNG rng;
       rng = cvRNG(cvGetTickCount());
       for (int i = 1; i < 6; i++)
           path.push_back(Point2f(i * 800 / 6.0, cvRandInt(&rng) % 800));

       Mat img(800, 800, CV_8UC3);
       img = 0;
       for (int i = 0; i < path.size(); i++)
           circle(img, path[i], 3, Scalar(0, 0, 255), 3); //BGR
       vector<Point2f> bezierPath = bezierCurve(path);
       for (int i = 0; i < bezierPath.size(); i++) {
           //circle(img, bezierPath[i], 3, Scalar(0, 255, 255), 3); //BGR
           img.at<cv::Vec3b>(cvRound(bezierPath[i].y), cvRound(bezierPath[i].x)) = { 0, 255, 255 };
       }
       imshow("black", img);
       if (waitKey(0) == 'q')
           break;
   }
   return 0;
}
vector<Point2f> bezierCurve(vector<Point2f> src)
{
   if (src.size() < 1)//這種情況是不允許出現的，出現只能證明程序出錯了
       return src;
   const float step = 0.01;//採集100個點，即1.0/step
   vector<Point2f> res;
   if (src.size() == 1) {//遞歸結束條件，k=0
       for (float t = 0; t < 1; t += step)
           res.push_back(src[0]);//爲了和其他情況保持一致，生成了1.0/step個一樣的點
       return res;
   }
   vector<Point2f> src1;
   vector<Point2f> src2;
   src1.assign(src.begin(), src.end() - 1);//分成兩部分，即Pi和Pi+1
   src2.assign(src.begin() + 1, src.end());
   for (int i = 0; i < src1.size(); i++)
       cout << src1[i] << endl;
   cout << endl;
   for (int i = 0; i < src2.size(); i++)
       cout << src2[i] << endl;
   cout << endl;
   vector<Point2f> pln1 = bezierCurve(src1);
   vector<Point2f> pln2 = bezierCurve(src2);
   for (float t = 0; t < 1; t += step) {
       Point2f temp;
       temp = (1.0 - t) * pln1[cvRound(1.0 / step * t)] + t * pln2[cvRound(1.0 / step * t)];
       res.push_back(temp);
   }
   return res;
}