《坐标转换代码改进》

1、验证示例

输入图形A的中点坐标（2，2）
在图形B中应该是（8，2）

2、代码

#include <stdio.h>
#include <math.h>

typedef struct {
    float x;
    float y;
} Point;

float dotProduct(Point p1, Point p2) {
    return p1.x * p2.x + p1.y * p2.y;
}

int isPointInRectangle(Point p, Point A, Point B, Point C, Point D) {
    Point AB = { B.x - A.x, B.y - A.y };
    Point BC = { C.x - B.x, C.y - B.y };
    Point CD = { D.x - C.x, D.y - C.y };
    Point DA = { A.x - D.x, A.y - D.y };
    Point AP = { p.x - A.x, p.y - A.y };
    Point BP = { p.x - B.x, p.y - B.y };
    Point CP = { p.x - C.x, p.y - C.y };
    Point DP = { p.x - D.x, p.y - D.y };

    return dotProduct(AB, AP) >= 0 && dotProduct(BC, BP) >= 0 &&
        dotProduct(CD, CP) >= 0 && dotProduct(DA, DP) >= 0;
}

Point mapPoint(Point p, Point A[], Point B[]) {
    Point A1 = A[0];
    Point B1 = B[0];
    Point A2 = A[2];
    Point B2 = B[2];

    // 计算矩形1的旋转角度
    float theta = atan2(A2.y - A1.y, A2.x - A1.x);
    // 计算矩形2的旋转角度
    float phi = atan2(B2.y - B1.y, B2.x - B1.x);
    // 计算总旋转角度
    float totalRotation = phi - theta;

    // 计算映射点相对于矩形1左上角的坐标
    float translatedX = p.x - A1.x;
    float translatedY = p.y - A1.y;

    // 应用旋转变换
    float rotatedX = translatedX * cos(totalRotation) - translatedY * sin(totalRotation);
    float rotatedY = translatedX * sin(totalRotation) + translatedY * cos(totalRotation);

    // 计算矩形2的缩放比例
    float scaleX = hypot(B2.x - B1.x, B2.y - B1.y) / hypot(A2.x - A1.x, A2.y - A1.y);
    float scaleY = hypot(B2.x - B1.x, B2.y - B1.y) / hypot(A2.x - A1.x, A2.y - A1.y);

    // 应用缩放变换
    rotatedX *= scaleX;
    rotatedY *= scaleY;

    // 计算映射点相对于矩形2左上角的坐标
    float finalX = B1.x + rotatedX;
    float finalY = B1.y + rotatedY;

    Point mappedPoint;
    mappedPoint.x = finalX;
    mappedPoint.y = finalY;

    return mappedPoint;
}

int main() {
    Point A1 = { 0, 0 };
    Point A2 = { 0, 4 };
    Point A3 = { 4, 4 };
    Point A4 = { 4, 0 };

    //Point B1 = { 0, 0 };
    //Point B2 = { sqrt(18), sqrt(18) };
    //Point B3 = { 2*sqrt(18), 0 };
    //Point B4 = { sqrt(18), -sqrt(18) };

    Point B1 = { 8, 0 };
    Point B2 = { 6, 2 };
    Point B3 = { 8, 4 };
    Point B4 = { 10, 2 };

    Point A[4] = { A1, A2, A3, A4 };
    Point B[4] = { B1, B2, B3, B4 };

    Point p;
    printf("请输入检测点的坐标 (x y): ");
    scanf("%f %f", &p.x, &p.y);

    if (isPointInRectangle(p, A1, A2, A3, A4)) {
        Point mappedPoint = mapPoint(p, A, B);
        printf("映射后的坐标：(%f, %f)\n", mappedPoint.x, mappedPoint.y);
    }
    else {
        printf("点不在矩形1内部。\n");
    }

    return 0;
}

3、实验结果

结果正确，但是没有只是验证了中点是正确的感觉说服力度不够，图形也没有啥其他特殊点能验证。目前也一个图形中只有中点我们可以人为计算出他映射后的结果，然后代码验证。