已知5个点,求舍弃误差最大一个的点,求最佳的圆心,用C++回答
时间: 2024-03-09 08:50:42 浏览: 16
为了找到舍弃误差最大的点和最佳的圆心,可以按照以下步骤进行:
1. 定义一个结构体表示点的坐标,包括 x 和 y 两个成员变量。
```c++
struct Point {
double x;
double y;
};
```
2. 定义一个函数用于计算圆心和半径,这里使用最小二乘法来拟合圆形。
```c++
void fitCircle(const std::vector<Point>& points, Point& center, double& radius) {
double sumX = 0.0, sumY = 0.0, sumX2 = 0.0, sumY2 = 0.0, sumXY = 0.0;
double sumX3 = 0.0, sumY3 = 0.0, sumX2Y = 0.0, sumXY2 = 0.0;
for (auto& p : points) {
double x = p.x;
double y = p.y;
sumX += x;
sumY += y;
sumX2 += x * x;
sumY2 += y * y;
sumXY += x * y;
sumX3 += x * x * x;
sumY3 += y * y * y;
sumX2Y += x * x * y;
sumXY2 += x * y * y;
}
double N = points.size();
double C = N * sumX2 - sumX * sumX + N * sumY2 - sumY * sumY;
double D = N * sumX3 - 2 * sumX * sumX2 + N * sumXY2 - sumY * sumXY;
double E = N * sumY2 - sumY * sumY + N * sumX2 - sumX * sumX;
double F = N * sumX2Y - sumX * sumY2 + N * sumXY - sumX * sumY;
double A = (E * D - F * C) / (D * D - C * E);
double B = (F * D - C * A) / (D * D - C * E);
center.x = A / (-2);
center.y = B / (-2);
radius = sqrt(A * A + B * B - 4 * C) / (2 * fabs(C));
}
```
3. 定义一个函数用于计算每个点到圆心的距离,并返回距离最大的点的下标。
```c++
int findMaxErrorPoint(const std::vector<Point>& points, const Point& center) {
int maxIndex = 0;
double maxError = 0.0;
for (int i = 0; i < points.size(); ++i) {
double error = fabs(sqrt((points[i].x - center.x) * (points[i].x - center.x) + (points[i].y - center.y) * (points[i].y - center.y)) - radius);
if (error > maxError) {
maxIndex = i;
maxError = error;
}
}
return maxIndex;
}
```
4. 最后,将上述三个函数结合起来,可以得到如下的代码:
```c++
int main() {
std::vector<Point> points = { {1.0, 2.0}, {3.0, 4.0}, {5.0, 6.0}, {7.0, 8.0}, {9.0, 10.0} };
Point center;
double radius;
fitCircle(points, center, radius);
int maxIndex = findMaxErrorPoint(points, center);
std::cout << "舍弃误差最大的点是 (" << points[maxIndex].x << ", " << points[maxIndex].y << ")" << std::endl;
std::cout << "最佳的圆心是 (" << center.x << ", " << center.y << ")" << std::endl;
return 0;
}
```