英文:
Distance Between a point to a line formed by two points
问题
我试图找到从点3（newPoint）到由点1（prevPoint）和点2（curPoint）形成的直线的垂直距离在C++中。
我之前使用了这个公式。但现在在一些交叉检查后，我对我的程序的正确性产生了疑问。
编辑：
所有点（prevPoint、curPoint和newPoint）都是Point类型的。
struct Point {string name;int x;int y;bool visited;};double d = distance(prevPoint, curPoint);double dx = (curPoint.x - prevPoint.x) / d;double dy = (curPoint.y - prevPoint.y) / d;double distToLine = abs((newPoint.x - prevPoint.x) * dy - (newPoint.y - prevPoint.y) * dx);
distance 函数：
double distance(Point a, Point b) {double dx = a.x - b.x;double dy = a.y - b.y;return sqrt(dx * dx + dy * dy);}
我的代码正确吗？如果不正确，我需要一些正确的公式。谢谢！
英文:
I am trying to find the perpendicular distance from point 3 (newPoint) to a line formed by point 1 (prevPoint) and point 2 (curPoint) in C++.
I was using this formula previously. But now I am questioning the correctness of my program upon some cross-checking.
EDIT:
all the points(prevPoint, curPoint and newPoint) is of Point type
struct Point {string name;int x;int y;bool visited;};double d = distance(prevPoint, curPoint);double dx = (curPoint.x - prevPoint.x) / d;double dy = (curPoint.y - prevPoint.y) / d;double distToLine = abs((newPoint.x - prevPoint.x) * dy - (newPoint.y - prevPoint.y) * dx);
The distance function:
double distance(Point a, Point b) {double dx = a.x - b.x;double dy = a.y - b.y;return sqrt(dx * dx + dy * dy);}
Is my code correct? I need some correct formula if its not. Thanks!
答案1
得分: 3
以下是您要翻译的内容：
double d = distance(prevPoint, curPoint);double area = abs((curPoint.x - prevPoint.x) * (newPoint.y - prevPoint.y) - (curPoint.y - prevPoint.y) * (newPoint.x - prevPoint.x));double distToLine = area / d;
英文:
The formula would be as follows in your case:
double d = distance(prevPoint, curPoint);double area = abs((curPoint.x - prevPoint.x) * (newPoint.y - prevPoint.y) - (curPoint.y - prevPoint.y) * (newPoint.x - prevPoint.x));double distToLine = area / d;
答案2
得分: 0
在以下测试代码中，你的代码与我的代码产生了相同的结果。
struct Point{int x, y;Point(int x = 0, int y = 0) : x(x), y(y) {}};double MyMethod(Point prevPoint, Point curPoint, Point newPoint){// 为简单起见，解决问题时使用以prevPoint为原点的坐标系。Point B(curPoint.x - prevPoint.x, curPoint.y - prevPoint.y);Point C(newPoint.x - prevPoint.x, newPoint.y - prevPoint.y);// 将矢量P表示为// P = t * B - C// 其中，t是标量。//// 我们需要做的是找到最小化此P长度的t值。// 目标函数是：// Err(t) = (P点积 P) = (t * Bx - Cx)^2 + (t * By - Cy)^2// 然后// dErr(t)/dt = 2 * (t * Bx - Cx) * Bx + 2 * (t * By - Cy) * By = 0// 因此// t = (CxBx + CyBy) / (BxBx + ByBy)double t = (C.x * B.x + C.y * B.y) / double(B.x * B.x + B.y * B.y);// 现在我们要获得的距离是P的长度（使用计算出的t值）。double Px = t * B.x - C.x;double Py = t * B.y - C.y;return sqrt(Px * Px + Py * Py);}namespace YourCode{double distance(Point a, Point b){double dx = a.x - b.x;double dy = a.y - b.y;return sqrt(dx * dx + dy * dy);}double YourMethod(Point prevPoint, Point curPoint, Point newPoint){double d = distance(prevPoint, curPoint);double dx = (curPoint.x - prevPoint.x) / d;double dy = (curPoint.y - prevPoint.y) / d;double distToLine = abs((newPoint.x - prevPoint.x) * dy - (newPoint.y - prevPoint.y) * dx);return distToLine;}}int main(int argc, char *argv[]){Point A(3, 10);Point B(-5, 22);Point C(1, 4);std::cout << YourCode::YourMethod(A, B, C) << std::endl;std::cout << MyMethod(A, B, C) << std::endl;return 0;}
英文:
In the following test code, your code had the same results as mine.
struct Point{int x,y;Point( int x=0, int y=0 ) : x(x),y(y) {}};double MyMethod( Point prevPoint, Point curPoint, Point newPoint ){//For simplicity, solve the problem in a coordinate system with the origin at prevPoint.Point B( curPoint.x-prevPoint.x, curPoint.y-prevPoint.y );Point C( newPoint.x-prevPoint.x, newPoint.y-prevPoint.y );// Considering the Vector P as// P = t*B - C// where, t is scalar.//// What we need to do is find the value of t that minimizes the length of this P.// Objective Function is :// Err(t) = (P dot P) = (t*Bx - Cx)^2 + (t*By - Cy)^2// And Then// dErr(t)/dt = 2*(t*Bx - Cx)Bx + 2*(t*By - Cy)By = 0// So// t = (CxBx + CyBy) / (BxBx + ByBy)double t = ( C.x*B.x + C.y*B.y ) / double(B.x*B.x + B.y*B.y);// Now the distance we want to obtain is length of P (with the calculated t value).double Px = t*B.x - C.x;double Py = t*B.y - C.y;return sqrt( Px*Px + Py*Py );}namespace YourCode{double distance(Point a, Point b){double dx = a.x - b.x;double dy = a.y - b.y;return sqrt(dx * dx + dy * dy);}double YourMethod( Point prevPoint, Point curPoint, Point newPoint ){double d = distance(prevPoint, curPoint);double dx = (curPoint.x - prevPoint.x) / d;double dy = (curPoint.y - prevPoint.y) / d;double distToLine = abs((newPoint.x - prevPoint.x) * dy - (newPoint.y - prevPoint.y) * dx);return distToLine;}}int main(int argc, char *argv[]){Point A( 3,10 );Point B( -5, 22 );Point C( 1,4 );std::cout << YourCode::YourMethod(A,B,C) << std::endl;std::cout << MyMethod(A,B,C) << std::endl;return 0;}
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