赞
踩
- #include <opencv2/opencv.hpp>
- #include <iostream>
- #include <vector>
- #include <cmath>
-
- using namespace std;
- using namespace cv;
-
- vector<Point2d> orig_point;
- vector<Point2d> control_point;
- double u1 = 0.3;//这两个参数取决于中间两个点的位置
- double u2 = 0.7;
-
- int main() {
-
- //1.四个贝塞尔曲线上的点
- orig_point.push_back(Point2d(0, 0));
- orig_point.push_back(Point2d(3, 5));
- orig_point.push_back(Point2d(7, 5));
- orig_point.push_back(Point2d(10, 0));
- Mat P(4, 2, CV_64F);
-
- //将点赋值给4*2的矩阵
- for (int i = 0; i < 4; i++) {
- P.at<double>(i, 0) = orig_point[i].x;
- P.at<double>(i, 1) = orig_point[i].y;
- }
- cout << "贝塞尔曲线上的四个点分别为:\n" << P << endl;
-
- //2.定义一个4*4的U矩阵,每一行对应的是四个贝塞尔曲线上的点参数
- Mat U = (Mat_ <double>(4, 4) <<
- 0, 0, 0, 1, //对应第一个点,此时u=0
- u1 * u1 * u1, u1 * u1, u1, 1, //对应第二个点,(u1^3,u1^2,u1,1)
- u2 * u2 * u2, u2 * u2, u2, 1,
- 1, 1, 1, 1);//对应第四个点,此时u=1
-
- //3.定义好M矩阵,这个是固定的
- Mat M = (Mat_ <double>(4, 4) <<
- -1, 3, -3, 1,
- 3, -6, 3, 0,
- -3, 3, 0, 0,
- 1, 0, 0, 0);
- //4.求U*M的逆
- Mat UM = U * M;
- //cout << UM.size() << endl;
- //cout << UM << endl;
- Mat UM_inv = UM.inv();
-
- //5.控制点,因为P=U*M*C,所以控制点C=UM的逆*P
- Mat C;
- C = UM_inv * P;
- cout << endl << "求得的控制点为:\n" << C << endl;
- }
-
结果验证,首尾两点坐标一定相同
2021年9月22日 -- add
Copyright © 2003-2013 www.wpsshop.cn 版权所有,并保留所有权利。