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我尝试使用 3D，但我是初学者，所以我首先尝试使用 2D 并且 z = 0 的值。我有一个点数组，其值在数组点 [] 中使用 std::vector 随机。我有函数 Distance(...) 和 CaculateF(...) 来计算 points[] 的新值并存储在数组 pnew[] 中。我需要绘制点 [] 并将它们移动到 pnew [] 的值，但我只知道先在数组点 [] 中绘制随机点，我不能将它们完全移动到 pnew [] 中的值。有谁能够帮我？！

#include<stdlib.h>
#include<glut.h>
#include<iostream>
#include<conio.h>
#include<math.h>
#include<omp.h>
#include<time.h>
#include<Windows.h>
#include<vector>

using namespace std;

struct Point
{
    float x, y , z;
    float vx, vy, vz; 
    unsigned long m;
    unsigned char r, g, b, a;
};
vector< Point > points, pnew;

void animation_points( int value )
{
    // move all points left each frame
    for( size_t i = 0; i < points.size(); ++i )
    {
        points[i].x -= 1;
        // wrap point around if it's moved off
        // the edge of our 100x100 area
        if( points[i].x < -50 )
        {
            points[i].x = 100 + points[i].x;
        }
    }

    glutPostRedisplay();
    glutTimerFunc(30, animation_points, 1);
}

void display(void)
{
    glClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT);

    glMatrixMode(GL_PROJECTION);
    glLoadIdentity();
    glOrtho(-50, 50, -50, 50, -1, 1);

    glMatrixMode(GL_MODELVIEW);
    glLoadIdentity();

    // draw
    glColor3ub( 255, 255, 255 );
    glEnableClientState( GL_VERTEX_ARRAY );
    glEnableClientState( GL_COLOR_ARRAY );
    glVertexPointer( 2, GL_FLOAT, sizeof(Point), &points[0].x );
    glColorPointer( 4, GL_UNSIGNED_BYTE, sizeof(Point), &points[0].r );
    glPointSize( 3.0 );
    glDrawArrays( GL_POINTS, 0, points.size() );
    glDisableClientState( GL_VERTEX_ARRAY );
    glDisableClientState( GL_COLOR_ARRAY );

    glFlush();
    glutSwapBuffers();
}

void reshape(int w, int h)
{
    glViewport(0, 0, w, h);
}

//Distance between i and j
float Distance(float x1,float y1, float z1, float x2, float y2, float z2)
{
    return (sqrt(pow(x1-x2,2) + pow(y1-y2,2) + pow(z1-z2,2)));
}

//Value of F on Ox, Oy, Oz
Point CalculateF(double d, Point a, Point b, int dt)
{
    Point F;
    float vnewx, vnewy, vnewz, xnew , ynew, znew;
    float G = 6.6742*pow(10,-11);
    float Fx = (G*a.m*b.m/pow(d,2)*(a.x-b.x)/d);
    float Fy = (G*a.m*b.m/pow(d,2)*(a.y-b.y)/d);
    float Fz = (G*a.m*b.m/pow(d,2)*(a.z-b.z)/d);

    vnewx = a.vx + Fx*dt/a.m;
    vnewy = a.vy + Fy*dt/a.m;
    vnewz = a.vz + Fz*dt/a.m;

    xnew = a.x + a.x*dt;
    ynew = a.y + a.y*dt;
    znew = a.z + a.z*dt;

    F.x = xnew;
    F.y = ynew;
    F.z = znew;
    F.vx = vnewx;
    F.vy = vnewy;
    F.vz = vnewz;
    F.m = a.m;
    return F;
}

int main(int argc, char **argv)
{
    // begin
    glutInit(&argc, argv);
    glutInitDisplayMode(GLUT_RGBA | GLUT_DEPTH | GLUT_DOUBLE);

    glutInitWindowSize(640,480);
    glutCreateWindow("N - body");

    glutDisplayFunc(display);
    glutReshapeFunc(reshape);
    // animation points
    //glutTimerFunc(30, animation_points, 1);

    ////
    int n, t;
    cout<<"\n Number of body: ";
    cin>>n;
// move after time t
    cout<<"\n\n Time: ";
    cin>>t;
    t *= 3600;

    // random points
    for( int i = 0; i < n; ++i )
    {
    Point pt;
    pt.x = -50 + (rand() % 100);
    pt.y = -50 + (rand() % 100);
    pt.z = 0;
    pt.r = rand() % 255;
    pt.g = rand() % 255;
    pt.b = rand() % 255;
    pt.a = 255;
    points.push_back(pt);
    }    

glutMainLoop();

float d;

//#pragma omp parallel default(shared) private(i,j)
for (int i = 0 ; i < n ; i++)
{
    //#pragma omp for schedule(static)
    for (int j = 0 ; j < n ; j++)
    {
        d = Distance(points[i].x, points[i].y,points[i].z, points[j].x, points[j].y, points[j].z);
        if (d!=0)
            points[i] = CalculateF(d,points[i], points[j], t); 
    }
    pnew.push_back(points[i]);
}

return 0;
}
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1 回答 1

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您需要将点的初始位置和目标位置存储在数组中，然后在渲染代码中在它们之间进行插值。为此，您确定已经过去了多少时间，double lambda从时间计算 0.0 到 1.0 范围内的 a，然后在 position 处绘制点p_start + lambda * (p_target - p_start)。

于 2013-05-13T18:09:52.960 回答