C++ 在给定斜率的线上找到给定距离处的点（Find points at a given distance on a line of given slope）

news2024/11/23 14:13:18

给定二维点 p(x 0 , y 0 )的坐标。找到距离该点 L 的点，使得连接这些点所形成的线的斜率为M。

例子：
输入： p = (2, 1)
L = sqrt(2)
M = 1
输出：3, 2
1, 0
解释：
与源的距离为 sqrt(2) ，并具有所需的斜率m = 1。

输入： p = (1, 0)
L = 5
M = 0
输出： 6, 0
-4, 0

我们需要找到与给定点距离为 L 的两个点，它们位于斜率为 M 的直线上。
这个想法已在下面的帖子中介绍:
C++ https://blog.csdn.net/hefeng_aspnet/article/details/141320964

Java https://blog.csdn.net/hefeng_aspnet/article/details/141321133

Python https://blog.csdn.net/hefeng_aspnet/article/details/141321178

C# https://blog.csdn.net/hefeng_aspnet/article/details/141321206

Javascript https://blog.csdn.net/hefeng_aspnet/article/details/141321238

根据输入的斜率，该问题可以分为 3 类。

1、如果斜率为零，我们只需要调整源点的 x 坐标

2、如果斜率无限大，则需要调整 y 坐标

3、对于其他斜率值，我们可以使用以下方程来找到点

现在利用上述公式我们可以找到所需的点。

// C++ program to find the points on a line of
// slope M at distance L
#include <bits/stdc++.h>
using namespace std;

// structure to represent a co-ordinate
// point
struct Point {

float x, y;
Point() { x = y = 0; }
Point(float a, float b) { x = a, y = b; }
};

// Function to print pair of points at
// distance 'l' and having a slope 'm'
// from the source
void printPoints(Point source, float l, int m)
{
// m is the slope of line, and the
// required Point lies distance l
// away from the source Point
Point a, b;

// slope is 0
if (m == 0) {
a.x = source.x + l;
a.y = source.y;

b.x = source.x - l;
b.y = source.y;
}

// if slope is infinite
else if (m == std::numeric_limits<float>::max()) {
a.x = source.x;
a.y = source.y + l;

b.x = source.x;
b.y = source.y - l;
}
else {
float dx = (l / sqrt(1 + (m * m)));
float dy = m * dx;
a.x = source.x + dx;
a.y = source.y + dy;
b.x = source.x - dx;
b.y = source.y - dy;
}

// print the first Point
cout << a.x << ", " << a.y << endl;

// print the second Point
cout << b.x << ", " << b.y << endl;
}

// driver function
int main()
{
Point p(2, 1), q(1, 0);
printPoints(p, sqrt(2), 1);
cout << endl;
printPoints(q, 5, 0);
return 0;
}