C语言编程立体像对的空间前方交会（分别已知两像片的像对坐标x1,x2和其方位元素和内方位元素）的严密共线方程计算窗口程序

以下是计算C语言编程立体像对的空间前方交会的严密共线方程的窗口程序：

#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <conio.h>
#include <graphics.h>

#define PI 3.14159265358979323846

void main()
{
    float x1, y1, x2, y2, Xs1, Ys1, f1, k1, Xs2, Ys2, f2, k2, L1, L2, M1, M2, N1, N2, X, Y, Z;
    int gd = DETECT, gm;

    printf("Enter the coordinates of image 1 (x1, y1): ");
    scanf("%f %f", &x1, &y1);

    printf("Enter the coordinates of image 2 (x2, y2): ");
    scanf("%f %f", &x2, &y2);

    printf("Enter the orientation elements of image 1 (Xs1, Ys1, f1, k1): ");
    scanf("%f %f %f %f", &Xs1, &Ys1, &f1, &k1);

    printf("Enter the orientation elements of image 2 (Xs2, Ys2, f2, k2): ");
    scanf("%f %f %f %f", &Xs2, &Ys2, &f2, &k2);

    // Convert orientation elements from degrees to radians
    f1 = f1 * PI / 180;
    k1 = k1 * PI / 180;
    f2 = f2 * PI / 180;
    k2 = k2 * PI / 180;

    // Compute the direction cosines of the ray in image 1
    L1 = cos(f1) * cos(k1);
    M1 = cos(f1) * sin(k1);
    N1 = sin(f1);

    // Compute the direction cosines of the ray in image 2
    L2 = cos(f2) * cos(k2);
    M2 = cos(f2) * sin(k2);
    N2 = sin(f2);

    // Compute the coefficients of the linear equation system
    float a[3][3] = {
        {L1 - L2, M1 - M2, N1 - N2},
        {Xs2 - Xs1, Ys2 - Ys1, -1},
        {x2 - x1, y2 - y1, 0}
    };

    float b[3] = {Xs2 - Xs1, y1 - Ys1, x1 - Xs1};

    // Solve the linear equation system
    float det = a[0][0] * (a[1][1] * a[2][2] - a[1][2] * a[2][1]) -
                a[0][1] * (a[1][0] * a[2][2] - a[1][2] * a[2][0]) +
                a[0][2] * (a[1][0] * a[2][1] - a[1][1] * a[2][0]);

    if (det == 0) {
        printf("The linear equation system does not have a unique solution.\n");
        return;
    }

    float x = (b[0] * (a[1][1] * a[2][2] - a[1][2] * a[2][1]) -
               b[1] * (a[0][1] * a[2][2] - a[0][2] * a[2][1]) +
               b[2] * (a[0][1] * a[1][2] - a[0][2] * a[1][1])) / det;

    float y = (b[1] * (a[0][0] * a[2][2] - a[0][2] * a[2][0]) -
               b[0] * (a[1][0] * a[2][2] - a[1][2] * a[2][0]) +
               b[2] * (a[0][0] * a[1][2] - a[0][2] * a[1][0])) / det;

    float z = (b[2] * (a[0][0] * a[1][1] - a[0][1] * a[1][0]) -
               b[0] * (a[1][1] * a[2][2] - a[1][2] * a[2][1]) +
               b[1] * (a[0][1] * a[2][2] - a[0][2] * a[2][1])) / det;

    // Compute the 3D coordinates of the intersection point
    X = x + Xs1;
    Y = y + Ys1;
    Z = z + f1;

    // Display the result
    printf("The 3D coordinates of the intersection point are:\n");
    printf("X = %f, Y = %f, Z = %f\n", X, Y, Z);

    // Draw the 3D point on the screen
    initgraph(&gd, &gm, "");

    int x_screen = X * 10 + 320;
    int y_screen = Y * 10 + 240;

    circle(x_screen, y_screen, 5);

    getch();
    closegraph();
}

该程序会提示用户输入图像1和图像2的像对坐标和方位元素，然后计算它们的严密共线方程，并求出它们的交点的三维坐标。最后，程序会在屏幕上绘制该点的图像。