計(jì)算機(jī)圖形學(xué)關(guān)于任意直線的對稱變換

1、實(shí)驗(yàn)3：關(guān)于任意直線的對稱變換實(shí)驗(yàn)類型：驗(yàn)證、設(shè)計(jì)所需時(shí)間：3學(xué)時(shí)主要實(shí)驗(yàn)內(nèi)容及要求：對于任意直線的二維圖形對稱變化的實(shí)驗(yàn)，要求輸入的直線是任意直線，直線的端點(diǎn)只能由鍵盤輸入或者鼠標(biāo)拾取，要做對稱變換的圖形也是一個(gè)任意圖形（至少應(yīng)是一個(gè)任意多邊形）。對稱變換，先分析如何使用一系列簡單變換來構(gòu)造題目要求的復(fù)合變換。本體要實(shí)現(xiàn)的變換可以用如下一組變換組合來實(shí)現(xiàn)： 將直線任一點(diǎn)移至與坐標(biāo)原點(diǎn)重合 將平移后的直線繞原點(diǎn)旋轉(zhuǎn)至與某一坐標(biāo)軸重合 將題目要求的對稱變換轉(zhuǎn)為實(shí)現(xiàn)已知圖形關(guān)于上述坐標(biāo)軸的對稱變換 按逆序求上述、變換的逆變換 將上述矩陣依次相乘得到最終的復(fù)合變換矩陣 則某一多邊形關(guān)于任意直線的對稱

2、變換就轉(zhuǎn)變?yōu)閷⒃摱噙呅蔚母黜旤c(diǎn)與上述求得的復(fù)合變換進(jìn)行矩陣乘法，求得變換后的新多邊形的各個(gè)頂點(diǎn)坐標(biāo)。根據(jù)上述流程，編程實(shí)現(xiàn)，并測試程序功能。源代碼：#include<iostream>#include<gl/glut.h>using namespace std;void Initial(void)glClearColor(1.0f,1.0f,1.0f,1.0f);glMatrixMode(GL_PROJECTION);gluOrtho2D(0.0,200.0,0.0,150.0);class CPointpublic: int x;int y;CPoint()CPoin

3、t(int x1,int y1)x=x1; y=y1; static CPoint ZeroMoveToXY(CPoint p, CPoint XY);/原始坐標(biāo)向屏幕坐標(biāo)XY的平移static CPoint ToZero(CPoint p);/關(guān)于原點(diǎn)對稱 static CPoint XYMoveToZero(CPoint p, CPoint XY);/XY坐標(biāo)向屏幕坐標(biāo)的平移static CPoint Charge_AllLine(CPoint p, CPoint line_start, CPoint line_end);/關(guān)于Ax+By+C = 0對稱;CPoint CPoint:Zer

4、oMoveToXY(CPoint p, CPoint XY)/原始坐標(biāo)向屏幕坐標(biāo)XY的平移CPoint result;int change33 = 1,0,0,0,1,0,-XY.x,-XY.y,1;int p13 = 0,0,0;for (int j = 0;j < 3;j+)p1j=p.x * change0j + p.y * change1j + change2j;result.x = p10;result.y = p11;return result;CPoint CPoint:ToZero(CPoint p)/關(guān)于原點(diǎn)對稱CPoint result;int change33 =

5、-1,0,0,0,-1,0,0,0,1;int p13 = 0,0,0;for (int j = 0;j < 3;j+)p1j=p.x * change0j + p.y * change1j + change2j;result.x = p10;result.y = p11;return result;CPoint CPoint:XYMoveToZero(CPoint p, CPoint XY)/XY坐標(biāo)向屏幕坐標(biāo)的平移CPoint result;int change33 = 1,0,0,0,1,0,XY.x,XY.y,1;int p13 = 0,0,0;for (int j = 0;j

6、< 3;j+)p1j=p.x * change0j + p.y * change1j + change2j;result.x = p10;result.y = p11;return result;CPoint CPoint:Charge_AllLine(CPoint p, CPoint line_start, CPoint line_end)/關(guān)于Ax+By+C = 0對稱double A,B,C;if(line_start.x = line_end.x)A = 1;B = 0;C = -line_start.x;else if(line_start.y = line_end.y)A =

7、 0;B = 1;C = -line_start.y;else A = 1000/(line_end.x - line_start.x);B = -1000/(line_end.y - line_start.y);C = 1000*(-line_start.x/(line_end.x-line_start.x)+line_start.y/(line_end.y - line_start.y);CPoint result;float Y_f = (-A*B*p.x + A*A*p.y - B*C)/(B*B+A*A);float X_f = (-A*B*p.y + B*B*p.x - A*C)/

8、(B*B+A*A);CPoint xy,m1,m2;xy.x=X_f;xy.y = Y_f;m1 = CPoint:ZeroMoveToXY(p, xy);m2 = CPoint:ToZero(m1);result = XYMoveToZero(m2, xy);return result;void Display(void)glClear(GL_COLOR_BUFFER_BIT);glColor3f(1.0f,0.0f,0.0f);glBegin(GL_LINES);int n,x,y,i; cout<<"請輸入對稱圖形的頂點(diǎn)數(shù)："<<endl; c

9、in>>n; CPoint *p=new CPointn; for(i=0;i<n;i+) cout<<"請輸入第"<<i+1<<"個(gè)頂點(diǎn)坐標(biāo)"<<endl; cin>>x>>y; pi.x=x; pi.y=y; if(i=0|i=1) glVertex2i(pi.x,pi.y); else if(i=n-1) glVertex2i(pi-1.x,pi-1.y); glVertex2i(pi.x,pi.y); glVertex2i(pi.x,pi.y); glVer

10、tex2i(p0.x,p0.y); else glVertex2i(pi-1.x,pi-1.y); glVertex2i(pi.x,pi.y); cout<<"請輸入對稱直線的兩個(gè)頂點(diǎn)："<<endl;CPoint *p2=new CPoint2;for(i=0;i<2;i+)cout<<"請輸入第"<<i+1<<"個(gè)頂點(diǎn)坐標(biāo)"<<endl; cin>>x>>y; p2i.x=x; p2i.y=y;glVertex2i(p2i.x,p

11、2i.y); for(i=0;i<n;i+)pi=CPoint:Charge_AllLine(pi, p20, p21);if(i=0|i=1) glVertex2i(pi.x,pi.y); else if(i=n-1) glVertex2i(pi-1.x,pi-1.y); glVertex2i(pi.x,pi.y); glVertex2i(pi.x,pi.y); glVertex2i(p0.x,p0.y); else glVertex2i(pi-1.x,pi-1.y); glVertex2i(pi.x,pi.y); glEnd(); glFlush();int main(int argc,char*argv)glutInit(&argc,argv);glutInitDisplayMode(