【充满原题的胡策】训练2.26 T2（三维计算几何）

题目：

题解：

这道题目基本分为4个步骤：
旋转地平面（100pts）
投影
求凸包
求面积
这个投影的过程其实不错，我只写了30pts的

代码：

#include <cstdio>
#include <algorithm>
using namespace std;
const double INF=1e18;
const double eps=1e-9;
int dcmp(double x)
{
    if (x<=eps && x>=-eps) return 0;
    return (x>0)?1:-1;
}
struct po
{
    double x,y,z;
    po (double X=0,double Y=0,double Z=0){x=X; y=Y; z=Z;}
}d[50005],sta[50005],light;
int n,top;double ans;
bool operator <(const po &a,const po&b){return a.x<b.x||(a.x==b.x&&a.y<b.y);}
double cj(po x,po y){return x.x*y.y-x.y*y.x;}
po operator -(po x,po y){return po(x.x-y.x,x.y-y.y);}
void tb()
{
    sort(d+1,d+n+1);
    top=0;
    for (int i=1;i<=n;i++)
    {
        while (top>1 && dcmp(cj(sta[top]-sta[top-1],d[i]-sta[top-1]))<=0) top--;
        sta[++top]=d[i];     
    }
    int k=top;
    for (int i=n-1;i>=1;i--)
    {
        while (top>k && dcmp(cj(sta[top]-sta[top-1],d[i]-sta[top-1]))<=0) top--;
        sta[++top]=d[i];
    }
    if (n>1) top--;
}
po ty(po a,po b)
{
    double x,y,z;
    if(dcmp(a.x-b.x))x=a.x+a.z*(a.x-b.x)/(b.z-a.z);
    else x=a.x;
    if(dcmp(a.y-b.y))y=a.y+a.z*(a.y-b.y)/(b.z-a.z);
    else y=a.y;
    z=0;
    return po(x,y,z);
}
int main()
{
    freopen("shadow.in","r",stdin);
    freopen("shadow.out","w",stdout);
    double x,y,z;
    for(int i=1;i<=3;i++)scanf("%lf%lf%lf",&x,&y,&z);
    scanf("%lf%lf%lf",&x,&y,&z);
    light=po(x,y,z);
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%lf%lf%lf",&x,&y,&z);
        d[i]=ty(light,po(x,y,z)); 
    }
    tb();
    double ans=0;
    for(int i=3;i<=top;i++) ans+=cj(sta[i-1]-sta[1],sta[i]-sta[1])/2;
    if(ans<0) ans=-ans;
    printf("%0.2lf\n",ans);
}