【2011 Greater New York Regional 】Problem I :The Golden Ceiling

一道比较简单但是繁琐的三维计算几何，找错误找的我好心酸，没想到就把一个变量给写错了 = =；

题目的意思是求平面切长方体的截面面积+正方体顶部所遮盖的面积；

找出所有的切点，然后二维凸包一下直接算面积即可！

发个代码纪念一下！

代码：

 #include<cstdio>
 #include<cmath>
 #include<algorithm>
 #include<cstring>
 #define eps 1e-8
 using namespace std;

 inline int sig(double x){return (x>eps)-(x<-eps);}
 double w,l,hh,aa,bb,cc,dd;
 int num1,num2;
 struct point
 {
     double x,y,z;
     point(double x=,double y=,double z=):x(x),y(y),z(z) { }
     bool operator < (const point &t)const
     {
         if(sig(x-t.x)==)
         {
             return y<t.y;
         }
         else return x<t.x;
     }
     point operator+(const point&b)const{return point(x+b.x,y+b.y,z+b.z);}
     point operator-(const point&b)const{return point(x-b.x,y-b.y,z-b.z);}
     point operator*(double p){return point(x*p,y*p,z*p);}
     point operator/(double p){return point(x/p,y/p,z/p);}
 } ve[],tu[],vv[],tt[];

 void intersection(point a,point b)
 {
     double t=(aa*a.x+bb*a.y+cc*a.z+dd)/(aa*(a.x-b.x)+bb*(a.y-b.y)+cc*(a.z-b.z));
     if(t>=-eps&&t<=1.00000+eps)
     {
         point v=a+(b-a)*t;
         ve[num1++]=v;
         if(sig(v.z-hh)==)
             tu[num2++]=v;
     }
 }
 double lenth(point a){return sqrt(a.x*a.x+a.y*a.y+a.z*a.z);}
 point cross(point a,point b){return point(a.y*b.z-a.z*b.y,a.z*b.x-a.x*b.z,a.x*b.y-a.y*b.x);}
 double area(point a,point b,point c){return lenth(cross(b-a,c-a))/2.0;}
 double cross2(point a,point b){return a.x*b.y-a.y*b.x;}
 bool check(point a)
 {
     if(sig(aa*a.x+bb*a.y+cc*a.z+dd)<)
         return ;
     return ;
 }

 int convexhull(point *p,int n,point* ch)
 {
     sort(p,p+n);
     int m=;
     for(int i=;i<n;i++)
     {
         while(m>&&cross2(ch[m-]-ch[m-],p[i]-ch[m-])<=eps)m--;
         ch[m++]=p[i];
     }
     int k=m;
     for(int i=n-;i>=;i--)
     {
         while(m>k&&cross2(ch[m-]-ch[m-],p[i]-ch[m-])<=eps)m--;
         ch[m++]=p[i];
     }
     if(n>)m--;
     return m;
 }

 int main()
 {
     int t,ca;
     scanf("%d",&t);
     while(t--)
     {
         memset(tu,,sizeof tu);
         memset(ve,,sizeof ve);
         memset(vv,,sizeof vv);
         memset(tt,,sizeof tt);
         scanf("%d",&ca);
         printf("%d ",ca);
         scanf("%lf%lf%lf%lf%lf%lf%lf",&l,&w,&hh,&aa,&bb,&cc,&dd);
         num1=num2=;
         dd=-dd;
         point a(0.0,0.0,0.0);
         point b(l,0.0,0.0);
         point c(l,w,0.0);
         point d(0.0,w,0.0);
         point e(0.0,0.0,hh);
         point f(l,0.0,hh);
         point g(l,w,hh);
         point h(0.0,w,hh);
         intersection(a,b);
         intersection(b,c);
         intersection(c,d);
         intersection(d,a);
         intersection(a,e);
         intersection(b,f);
         intersection(c,g);
         intersection(d,h);
         intersection(e,f);
         intersection(f,g);
         intersection(g,h);
         intersection(h,e);
         if(check(e)==)tu[num2++]=e;
         if(check(f)==)tu[num2++]=f;
         if(check(g)==)tu[num2++]=g;
         if(check(h)==)tu[num2++]=h;
         int x=convexhull(ve,num1,vv);
         int y=convexhull(tu,num2,tt);
         double ans=;
         for(int i=;i<x;i++)
             ans+=area(vv[],vv[i-],vv[i]);
         for(int i=;i<y;i++)
             ans+=area(tt[],tt[i-],tt[i]);
         printf("%.0lf\n",ceil(ans));
     }
     return ;
 }