hdu4087ALetter to Programmers(三维旋转矩阵)

参考

三维旋转矩阵 + 矩阵加速

这个还要用到仿射变换。

平移

translate tx ty tz

1 0 0 tx

0 1 0 ty

0 0 1 tz

0 0 0 1

缩放

scale kx ky kz

kx 0  0  0

0  ky 0  0

0  0  kz 0

0  0  0  1

绕任意轴（过原点）旋转（注意要把轴向量归一化，不然会在“点在轴上”这个情况下出问题）

rotate x y z d

(1-cos(d))*x*x+cos(d)     (1-cos(d))*x*y-sin(d)*z   (1-cos(d))*x*z+sin(d)*y   0

(1-cos(d))*y*x+sin(d)*z   (1-cos(d))*y*y+cos(d)     (1-cos(d))*y*z-sin(d)*x   0

(1-cos(d))*z*x-sin(d)*y   (1-cos(d))*z*y+sin(d)*x   (1-cos(d))*z*z+cos(d)     0

                 0                                     0                                      0                       1

 

类似于一种构造矩阵的方式，每一种操作都相当于乘一次矩阵，具体第三个怎么推出来的没看懂。。。

 

矩阵乘法不支持交换律，一定要看好顺序。。debug了一天。。

据说会有-0.00这种答案不通过的坑，不过下面的代码加了eps..

 

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 1010
 #define LL long long
 #define INF 0xfffffff
 #define forn(i,n) for(int i=0; i<(int)(n); i++)
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;

 struct point3
 {
     double x,y,z;
 } p[N];
 struct Mat
 {
     double mat[][];
 };
 int n,m;
 char str[];
 Mat operator * (Mat a,Mat b)
 {
     Mat c;
     memset(c.mat,,sizeof(c.mat));
     int i,j,k;
     for(i =  ; i < n ; i++)
         for(j =  ; j < n ; j++)
             for(k = ; k < n ; k++)
                 c.mat[i][j] +=a.mat[i][k]*b.mat[k][j];
     return c;
 }
 Mat operator ^(Mat a,int k)
 {
     Mat c;

     int i,j;
     for(i = ; i < n ; i++)
         for(j = ; j < n ; j++)
             c.mat[i][j] = (i==j);
     for(; k ; k >>= )
     {
         if(k&) c = c*a;
         a = a*a;
     }
     return c;
 }
 Mat solve(int k)
 {

     Mat cc;
     double a[];
     int i;
     memset(cc.mat,,sizeof(cc.mat));
     for(i =  ; i <  ; i++) cc.mat[i][i] = ;
     while(~scanf("%s",str))
     {
         if(str[]=='e')
         {
             break;
         }
         Mat c;
         memset(c.mat,,sizeof(c.mat));
         for(i =  ; i <  ; i++) c.mat[i][i] = ;
         if(str[]=='t')
         {
             forn(i,) scanf("%lf", &a[i]);
             forn(i,) c.mat[i][]=a[i];
         }
         else if(str[]=='s')
         {
             for(i =  ; i <  ; i++)
                 scanf("%lf",&a[i]);
             for(i =  ; i < ; i++)
                 c.mat[i][i] = a[i];
         }
         else if(str[]=='r'&&str[]=='o')
         {
             for(i =  ; i < ; i++) scanf("%lf",&a[i]);
             a[] = a[]/*pi;
             double mag=sqrt(a[]*a[]+a[]*a[]+a[]*a[]);
             a[]/=mag;
             a[]/=mag;
             a[]/=mag;
             c.mat[][] = (-cos(a[]))*a[]*a[]+cos(a[]);
             c.mat[][] = (-cos(a[]))*a[]*a[]-sin(a[])*a[];
             c.mat[][] = (-cos(a[]))*a[]*a[]+sin(a[])*a[];
             c.mat[][] = (-cos(a[]))*a[]*a[]+sin(a[])*a[];
             c.mat[][] = (-cos(a[]))*a[]*a[]+cos(a[]);
             c.mat[][] = (-cos(a[]))*a[]*a[]-sin(a[])*a[];
             c.mat[][] = (-cos(a[]))*a[]*a[]-sin(a[])*a[];
             c.mat[][] = (-cos(a[]))*a[]*a[]+sin(a[])*a[];
             c.mat[][] = (-cos(a[]))*a[]*a[]+cos(a[]);
         }
         else if(str[]=='r')
         {
             int kk ;
             scanf("%d",&kk);
             c = solve(kk);
         }
         cc = c*cc;
     }
     return cc^k;
 }
 int dcmp(double x)
 {
     if(fabs(x)<eps) return ;
     return x<?-:;
 }
 int main()
 {
     int i,j;
     n = ;
     while(scanf("%d",&m)&&m)
     {
         Mat c;
         memset(c.mat,,sizeof(c.mat));
         for(i = ; i < n; i++) c.mat[i][i] = ;
         c = solve()*c;
         for(i = ; i <= m; i++)
         {
             Mat x,y;
             memset(y.mat,,sizeof(y.mat));
             for(j =  ; j <  ; j++)
                 scanf("%lf",&y.mat[j][]);
             y.mat[][] = ;
             x = c*y;
             p[i].x = x.mat[][],p[i].y = x.mat[][],p[i].z = x.mat[][];
         }
         for(i = ; i <= m; i++)
         {
             printf("%.2f %.2f %.2f\n",p[i].x+eps,p[i].y+eps,p[i].z+eps);
         }
         puts("");
     }
     return ;
 }