2014
03-16

# openGL

Jiaoshou selected a course about “openGL” this semester. He was quite interested in modelview, which is a part of “openGL”. Just using three functions, it could make the model to move, rotate and largen or lessen. But he was puzzled with the theory of the modelview. He didn’t know a vertex after several transformations where it will be.

Now, He tells you the position of the vertex and the transformations. Please help Jiaoshou find the position of the vertex after several transformations.

The input will start with a line giving the number of test cases, T.
Each case will always begin with “glBegin(GL_POINTS);”.Then the case will be followed by 5 kinds of function.
1. glTranslatef(x,y,z);
This function will translate the vertex(x’,y’,z’) to vertex(x+x’,y+y’,z+z’).
2. glRotatef(angle,x,y,z);
This function will turn angle radians counterclockwise around the axis (0,0,0)->(x,y,z).
3. glScalef(x,y,z);
This function wiil translate the vertex(x’,y’,z’) to vertex(x*x’,y*y’,z*z’).
4. glVertex3f(x,y,z);
This function will draw an initial vertex at the position(x,y,z). It will only appear once in one case just before “glEnd();”. In openGL, the transformation matrices are right multiplied by vertex matrix. So you should do the transformations in the reverse order.
5. glEnd();
This function tells you the end of the case.
In this problem angle,x,y,z are real numbers and range from -50.0 to 50.0. And the number of functions in each case will not exceed 100.

The input will start with a line giving the number of test cases, T.
Each case will always begin with “glBegin(GL_POINTS);”.Then the case will be followed by 5 kinds of function.
1. glTranslatef(x,y,z);
This function will translate the vertex(x’,y’,z’) to vertex(x+x’,y+y’,z+z’).
2. glRotatef(angle,x,y,z);
This function will turn angle radians counterclockwise around the axis (0,0,0)->(x,y,z).
3. glScalef(x,y,z);
This function wiil translate the vertex(x’,y’,z’) to vertex(x*x’,y*y’,z*z’).
4. glVertex3f(x,y,z);
This function will draw an initial vertex at the position(x,y,z). It will only appear once in one case just before “glEnd();”. In openGL, the transformation matrices are right multiplied by vertex matrix. So you should do the transformations in the reverse order.
5. glEnd();
This function tells you the end of the case.
In this problem angle,x,y,z are real numbers and range from -50.0 to 50.0. And the number of functions in each case will not exceed 100.

1
glBegin(GL_POINTS);
glScalef(2.0,0.5,3.0);
glTranslatef(0.0,1.0,0.0);
glVertex3f(1.0,1.0,1.0);
glEnd();

2.0 1.0 3.0
Hint
In this sample, we first let the vertex do “glTranslatef(x,y,z);” this function, then do “glScalef(x,y,z)”.


#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#define eps 1e-8
using namespace std;
const double PI=acos(-1.0);
const int MAXN=105;

struct point3{
double x,y,z;
};

struct order{
char c;
double x,y,z,ang;
}ord[MAXN];

point3 rotate3(point3 q,double x0,double y0,double z0,double ang)
{
double x2=x0*x0;
double y2=y0*y0;
double z2=z0*z0;
double d2=x2+y2+z2;
double d=sqrt(d2);
double sina=sin(ang);
double cosa=cos(ang);
point3 ans;
ans.x=(x2+(y2+z2)*cosa)/d2*q.x+(x0*y0*(1-cosa)/d2-z0*sina/d)*q.y+(x0*z0*(1-cosa)/d2+y0*sina/d)*q.z;
ans.y=(y0*x0*(1-cosa)/d2+z0*sina/d)*q.x+(y2+(x2+z2)*cosa)/d2*q.y+(y0*z0*(1-cosa)/d2-x0*sina/d)*q.z;
ans.z=(z0*x0*(1-cosa)/d2-y0*sina/d)*q.x+(z0*y0*(1-cosa)/d2+x0*sina/d)*q.y+(z2+(x2+y2)*cosa)/d2*q.z;
return ans;
}

int main()
{
int T;
char str[500];
scanf("%d",&T);
while(T--){
int i,j;
double tmp;
double tail,muti;
double sig;
for(i=0;;i++){
scanf("%s",str);
if(str[2]=='B') ord[i].c='B';
else if(str[2]=='S'){
ord[i].c='S';
for(j=2;str[j]!='(';j++);
tmp=0.0;
sig=1.0;
for(j++;str[j]!='.';j++){
if(str[j]=='-'){
sig=-1.0;
continue;
}
tmp=tmp*10+str[j]-'0';
}
tail=0.0;
muti=0.1;
for(j++;str[j]!=',';j++){
tail+=(str[j]-'0')*muti;
muti/=10.0;
}
tmp+=tail;
ord[i].x=tmp*sig;
tmp=0.0;
sig=1.0;
for(j++;str[j]!='.';j++){
if(str[j]=='-'){
sig=-1.0;
continue;
}
tmp=tmp*10+str[j]-'0';
}
tail=0.0;
muti=0.1;
for(j++;str[j]!=',';j++){
tail+=(str[j]-'0')*muti;
muti/=10.0;
}
tmp+=tail;
ord[i].y=tmp*sig;
tmp=0.0;
sig=1.0;
for(j++;str[j]!='.';j++){
if(str[j]=='-'){
sig=-1.0;
continue;
}
tmp=tmp*10+str[j]-'0';
}
tail=0.0;
muti=0.1;
for(j++;str[j]!=')';j++){
tail+=(str[j]-'0')*muti;
muti/=10.0;
}
tmp+=tail;
ord[i].z=tmp*sig;
}
else if(str[2]=='T'){
ord[i].c='T';
for(j=2;str[j]!='(';j++);
tmp=0.0;
sig=1.0;
for(j++;str[j]!='.';j++){
if(str[j]=='-'){
sig=-1.0;
continue;
}
tmp=tmp*10+str[j]-'0';
}
tail=0.0;
muti=0.1;
for(j++;str[j]!=',';j++){
tail+=(str[j]-'0')*muti;
muti/=10.0;
}
tmp+=tail;
ord[i].x=tmp*sig;
tmp=0.0;
sig=1.0;
for(j++;str[j]!='.';j++){
if(str[j]=='-'){
sig=-1.0;
continue;
}
tmp=tmp*10+str[j]-'0';
}
tail=0.0;
muti=0.1;
for(j++;str[j]!=',';j++){
tail+=(str[j]-'0')*muti;
muti/=10.0;
}
tmp+=tail;
ord[i].y=tmp*sig;
tmp=0.0;
sig=1.0;
for(j++;str[j]!='.';j++){
if(str[j]=='-'){
sig=-1.0;
continue;
}
tmp=tmp*10+str[j]-'0';
}
tail=0.0;
muti=0.1;
for(j++;str[j]!=')';j++){
tail+=(str[j]-'0')*muti;
muti/=10.0;
}
tmp+=tail;
ord[i].z=tmp*sig;
}
else if(str[2]=='V'){
ord[i].c='V';
for(j=2;str[j]!='(';j++);
tmp=0.0;
sig=1.0;
for(j++;str[j]!='.';j++){
if(str[j]=='-'){
sig=-1.0;
continue;
}
tmp=tmp*10+str[j]-'0';
}
tail=0.0;
muti=0.1;
for(j++;str[j]!=',';j++){
tail+=(str[j]-'0')*muti;
muti/=10.0;
}
tmp+=tail;
ord[i].x=tmp*sig;
tmp=0.0;
sig=1.0;
for(j++;str[j]!='.';j++){
if(str[j]=='-'){
sig=-1.0;
continue;
}
tmp=tmp*10+str[j]-'0';
}
tail=0.0;
muti=0.1;
for(j++;str[j]!=',';j++){
tail+=(str[j]-'0')*muti;
muti/=10.0;
}
tmp+=tail;
ord[i].y=tmp*sig;
tmp=0.0;
sig=1.0;
for(j++;str[j]!='.';j++){
if(str[j]=='-'){
sig=-1.0;
continue;
}
tmp=tmp*10+str[j]-'0';
}
tail=0.0;
muti=0.1;
for(j++;str[j]!=')';j++){
tail+=(str[j]-'0')*muti;
muti/=10.0;
}
tmp+=tail;
ord[i].z=tmp*sig;
}
else if(str[2]=='E') break;
else{
ord[i].c='R';
for(j=2;str[j]!='(';j++);
tmp=0.0;
sig=1.0;
for(j++;str[j]!='.';j++){
if(str[j]=='-'){
sig=-1.0;
continue;
}
tmp=tmp*10+str[j]-'0';
}
tail=0.0;
muti=0.1;
for(j++;str[j]!=',';j++){
tail+=(str[j]-'0')*muti;
muti/=10.0;
}
tmp+=tail;
ord[i].ang=tmp*sig;
tmp=0.0;
sig=1.0;
for(j++;str[j]!='.';j++){
if(str[j]=='-'){
sig=-1.0;
continue;
}
tmp=tmp*10+str[j]-'0';
}
tail=0.0;
muti=0.1;
for(j++;str[j]!=',';j++){
tail+=(str[j]-'0')*muti;
muti/=10.0;
}
tmp+=tail;
ord[i].x=tmp*sig;
tmp=0.0;
sig=1.0;
for(j++;str[j]!='.';j++){
if(str[j]=='-'){
sig=-1.0;
continue;
}
tmp=tmp*10+str[j]-'0';
}
tail=0.0;
muti=0.1;
for(j++;str[j]!=',';j++){
tail+=(str[j]-'0')*muti;
muti/=10.0;
}
tmp+=tail;
ord[i].y=tmp*sig;
tmp=0.0;
sig=1.0;
for(j++;str[j]!='.';j++){
if(str[j]=='-'){
sig=-1.0;
continue;
}
tmp=tmp*10+str[j]-'0';
}
tail=0.0;
muti=0.1;
for(j++;str[j]!=')';j++){
tail+=(str[j]-'0')*muti;
muti/=10.0;
}
tmp+=tail;
ord[i].z=tmp*sig;
}
}
point3 pnt;
i--;
pnt.x=ord[i].x;
pnt.y=ord[i].y;
pnt.z=ord[i].z;
for(i--;i>=0;i--){
if(ord[i].c=='T'){
pnt.x+=ord[i].x;
pnt.y+=ord[i].y;
pnt.z+=ord[i].z;
}
else if(ord[i].c=='S'){
pnt.x*=ord[i].x;
pnt.y*=ord[i].y;
pnt.z*=ord[i].z;
}
else if(ord[i].c=='R'){
pnt=rotate3(pnt,ord[i].x,ord[i].y,ord[i].z,ord[i].ang);
}
}
printf("%.1f %.1f %.1f\n",pnt.x,pnt.y,pnt.z);
}
return 0;
}

1. 这道题目的核心一句话是：取还是不取。
如果当前取，则index+1作为参数。如果当前不取，则任用index作为参数。