1502: [NOI2005]月下柠檬树

题目链接

题目大意：有一个由圆锥和圆台组成的柠檬树，在月亮发出的平行光下，可以形成一个影子，求这个影子的面积

题解：自适应辛普森积分……

我的收获：2333

#include
#include
#include
#include
#include
using namespace std;
const double eps=1e-5;
struct circle
{
     double x,y;
     double r;
}a[10001];
struct circle2
{
     double x,y;
     double r;
     double ll,rr;
}sx[10001];
struct line
{
     double x1,y1,x2,y2;
}lin[10001];
double h[10001];
bool flag[10001];
int p,n;
inline double fabs(double x)
{
     if(x<0)
          x=-x;
     return x;
}
inline double max(double x,double y)
{
     if(x>y)
          return x;
     return y;
}
inline double geth(double x)
{
     int i;
     double len=0,dis;
     for(i=1;i<=n;i++)
     {
          if(sx[i].ll>=x||sx[i].rr<=x)
               continue;
          dis=sqrt(sx[i].r*sx[i].r-(sx[i].x-x)*(sx[i].x-x));
          len=max(len,dis*(double)2);
     }
     for(i=1;i<=p;i++)
     {
          if(lin[i].x1>x||lin[i].x2continue;
          dis=(lin[i].y1+(lin[i].y2-lin[i].y1)/(lin[i].x2-lin[i].x1)*(x-lin[i].x1));
          len=max(len,dis*(double)2);
     }
     return len;
}
inline double cal(double len,double lh,double mh,double rh)
{
     return (lh+mh*(double)4+rh)*len/(double)6;
}
inline double simpson(double l,double mid,double r,double lh,double mh,double rh,double s)
{
     double mid1=(l+mid)/(double)2,mid2=(mid+r)/(double)2;
     double mh1=geth(mid1),mh2=geth(mid2);
     double s1=cal(mid-l,lh,mh1,mh),s2=cal(r-mid,mh,mh2,rh);
     if(fabs(s1+s2-s)return s1+s2;
     return simpson(l,mid1,mid,lh,mh1,mh,s1)+simpson(mid,mid2,r,mh,mh2,rh,s2);
}
int main()
{
     memset(flag,false,sizeof(flag));
     double alpha;
     scanf("%d%lf",&n,&alpha);
     int i,j;
     for(i=0;i<=n;i++)
     {
          scanf("%lf",&h[i]);
          if(i!=0)
               h[i]=h[i]+h[i-1];
          a[i+1].x=h[i]/tan(alpha);
          a[i+1].y=0;
     }
     double x;
     for(i=1;i<=n;i++)
     {
          scanf("%lf",&x);
          a[i].r=x;
     }
     for(i=1;i<=n;i++)
     {
          if(a[i+1].x-a[i].x<fabs(a[i+1].r-a[i].r))
               continue;
          p++;
          lin[p].x1=a[i].x-a[i].r*(a[i+1].r-a[i].r)/(a[i+1].x-a[i].x);
          lin[p].y1=a[i].y+sqrt(a[i].r*a[i].r-(a[i].x-lin[p].x1)*(a[i].x-lin[p].x1));
          lin[p].x2=a[i+1].x-a[i+1].r*(a[i+1].r-a[i].r)/(a[i+1].x-a[i].x);
          lin[p].y2=a[i+1].y+sqrt(a[i+1].r*a[i+1].r-(a[i+1].x-lin[p].x2)*(a[i+1].x-lin[p].x2));
     }
     for(i=1;i<=n;i++)
     {
          sx[i].r=a[i].r;
          sx[i].x=a[i].x;
          sx[i].y=a[i].y;
          sx[i].ll=a[i].x-a[i].r;
          sx[i].rr=a[i].x+a[i].r;
     }
     double l=sx[1].x-sx[1].r,r=lin[p].x2,mid,lh,rh,mh;
     for(i=1;i<=n;i++)
     {
          l=min(l,sx[i].ll);
          r=max(r,sx[i].rr);
     }
     mid=(l+r)/(double)2;
     double ans=0;
     lh=geth(l);
     rh=geth(r);
     mh=geth(mid);
     ans=simpson(l,mid,r,lh,mh,rh,cal(r-l,lh,mh,rh));
     printf("%.2lf\n",ans);
     return 0;
}