用C语言求线段或直线与线段的交点
时间: 2023-09-15 07:15:14 浏览: 126
线段与线段的交点可以通过以下步骤求得:
1. 求出两条线段的参数方程,分别为 P1(t) = P1 + t * V1 和 P2(s) = P2 + s * V2,其中 P1 和 P2 分别是两条线段的起点,V1 和 V2 是它们的方向向量。
2. 将这两个参数方程联立得到一个二元一次方程组,即:
P1x + t * V1x = P2x + s * V2x
P1y + t * V1y = P2y + s * V2y
解这个方程组可以得到 t 和 s 的值。
3. 如果 0 <= t <= 1 且 0 <= s <= 1,则两条线段相交,交点为 P = P1 + t * V1。
以下是一个求线段与线段交点的示例代码:
```
#include <stdio.h>
typedef struct {
float x;
float y;
} Point;
typedef struct {
Point p1;
Point p2;
} Line;
Point getIntersection(Line l1, Line l2) {
float t, s;
float a = l1.p1.x - l2.p1.x;
float b = l1.p1.y - l2.p1.y;
float c = l2.p2.x - l2.p1.x;
float d = l1.p2.x - l1.p1.x;
float e = l2.p2.y - l2.p1.y;
float f = l1.p2.y - l1.p1.y;
float denom = a * e - b * d;
Point intersection = {0, 0};
if (denom == 0) {
// Lines are parallel
return intersection;
}
t = (c * b - a * e) / denom;
s = (d * b - a * f) / denom;
if (t >= 0 && t <= 1 && s >= 0 && s <= 1) {
intersection.x = l1.p1.x + t * (l1.p2.x - l1.p1.x);
intersection.y = l1.p1.y + t * (l1.p2.y - l1.p1.y);
}
return intersection;
}
int main() {
Line l1 = {{1, 1}, {3, 3}};
Line l2 = {{1, 3}, {3, 1}};
Point intersection = getIntersection(l1, l2);
printf("Intersection: (%f, %f)\n", intersection.x, intersection.y);
return 0;
}
```