sgu265：Wizards(计算几何)

题目大意：
$~~~~~~$一个空间直角坐标系内，对一个点有三种操作：
$~~~~~~$$1.$平移$(x,y,z)\Rightarrow (x+a,y+b,z+c)$；
$~~~~~~$$2.$缩放$(x,y,z)\Rightarrow (ax,by,cz)$；
$~~~~~~$$3.$绕一个向量$(x,y,z)$顺时针旋转$\alpha$角$($角度制$)$。
$~~~~~~$对空间内的$m$个点给出$n$个操作，求每个点在操作后的坐标。

分析：
$~~~~~~$我们用$\vec{i},\vec{j},\vec{k}$表示$x,y,z$轴上的单位向量，和一个记录平移的$\vec{p}$向量$($根据二维的仿射变换可以知道平移是不影响缩放和旋转的$)$。
$~~~~~~$平移直接给$\vec{p}$平移即可；
$~~~~~~$缩放对$\vec{p},\vec{i},\vec{j},\vec{k}$依次缩放；
$~~~~~~$旋转对$\vec{p},\vec{i},\vec{j},\vec{k}$依次旋转；
$~~~~~~$对于每个点$(x,y,z)$，操作后就是$(x\vec{i},y\vec{j},z\vec{k})+\vec{p}$。

$~~~~~~$关键就是怎样旋转。
$~~~~~~$对于点$p$绕$\vec{q}$旋转$\alpha$，求出$\vec{p}$在$\vec{q}$上的投影$\vec{t}$，令$\vec{p}=\vec{p}-\vec{t}$，那么$\vec{m}=\vec{q}\times \vec{p}$与$\vec{p}$构成的平面与$\vec{q}$垂直，将$\vec{m}$调整为$\vec{p}$的长度，旋转后的点即为 $\cos(\alpha)\vec{p}+\sin(\alpha)\vec{m}+\vec{t}$.

分析：

#include <cstdio>
#include <cmath>
typedef double DB;
using namespace std;

const DB PI = acos(-1.0);
const DB eps = 1e-10;

struct pot
{
    DB x, y, z;
    pot(DB x = 0, DB y = 0, DB z = 0):x(x),y(y),z(z){}
    void read() {scanf("%lf%lf%lf", &x, &y, &z);}
    void print() {printf("%lf %lf %lf\n", x, y, z);}
    void trans(DB tx, DB ty, DB tz) {x += tx, y += ty, z += tz;}
    void scal(DB tx, DB ty, DB tz) {x *= tx, y *= ty, z *= tz;}
    void rotate(const pot &a, DB alpha)
    {
        pot p = *this, tmp;
        tmp = a*((a*p)/(a*a));
        p = p-tmp;pot m = a^p;
        if(m.len() < eps) return ; 
        m.adjust(p.len());
        *this = p*cos(alpha)+m*sin(alpha)+tmp;
    }
    friend pot operator + (const pot &a, const pot &b) {return pot(a.x+b.x, a.y+b.y, a.z+b.z);}
    friend pot operator - (const pot &a, const pot &b) {return pot(a.x-b.x, a.y-b.y, a.z-b.z);}
    friend pot operator * (const pot &a, DB k) {return pot(a.x*k, a.y*k, a.z*k);}
    friend pot operator / (const pot &a, DB k) {return pot(a.x/k, a.y/k, a.z/k);}
    friend DB operator * (const pot &a, const pot &b) {return a.x*b.x+a.y*b.y+a.z*b.z;}
    friend pot operator ^ (const pot &a, const pot &b) {return pot(a.y*b.z-a.z*b.y, a.z*b.x-a.x*b.z, a.x*b.y-a.y*b.x);}
    DB len() {return sqrt((*this)*(*this));}
    void adjust(DB l) {*this = *this*(l/len());}
}p[4], v;

int main()
{
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif

    int n, m;
    scanf("%d", &n);
    p[1].x = p[2].y = p[3].z = 1;
    while(n--)
    {
        DB x, y, z, a;
        char q[3] = "\0";
        scanf("%s", q);
        scanf("%lf%lf%lf", &x, &y, &z);
        if(q[0] == 'T') p[0].trans(x, y, z);
        else if(q[0] == 'S') 
            for(int i = 0; i < 4; ++i)
                p[i].scal(x, y, z); 
        else
        {
            scanf("%lf", &a);a *= PI/180;
            for(int i = 0; i < 4; ++i)
                p[i].rotate(pot(x, y, z), a);
        }
    }
    scanf("%d", &m);
    while(m--)
    {
        v.read();
        v = p[0]+p[1]*v.x+p[2]*v.y+p[3]*v.z;
        v.print();
    }

    #ifndef ONLINE_JUDGE
    fclose(stdin);
    fclose(stdout);
    #endif
    return 0;
}