求凸包

BZOJ2829

#include <cstdio>
#include <algorithm>
#include <cmath>
#define LDB long double
using namespace std;

  const LDB eps=1e-8;

  struct data{
      LDB x,y,alp;
  }tmp[1000001],a[1000001],sta[1000010];
  
  int n,cnt;
  LDB xmini,ymini,bb,aa,r,ang[4],stddis;
  
  inline data operator - (const data&a,const data&b) {return (data){a.x-b.x,a.y-b.y,a.alp};}
  
  LDB X(data a,data b){
      return(a.x*b.y-a.y*b.x);
  }

  LDB dis(data a,data b){
      return(sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)));
  }

  int mycomp(const data&aa,const data&b){
      if (aa.alp<b.alp-eps) return(1);
      if (aa.alp>b.alp+eps) return(0);
      return(dis(a[1],aa)>dis(a[1],b));
  }

  void makepoint(LDB x,LDB y,LDB ag){
      for (int i=0;i<4;i++){
        tmp[++cnt].x=x+stddis*cos(ang[i]+ag);
        tmp[cnt].y=y+stddis*sin(ang[i]+ag);
    }
  }
  
  int main(){
    freopen("a.in","r",stdin);
    freopen("wrong.out","w",stdout);

      scanf("%d",&n);
      scanf("%Lf%Lf%Lf",&bb,&aa,&r);aa-=2*r;bb-=2*r;
      ang[0]=atan2(bb,aa),ang[1]=atan2(bb,-aa),ang[2]=atan2(-bb,-aa),ang[3]=atan2(-bb,aa);
      stddis=sqrt(aa*aa/4+bb*bb/4);
      for (int i=1;i<=n;i++){
        LDB t1,t2,t3;
        scanf("%Lf%Lf%Lf",&t1,&t2,&t3);
      makepoint(t1,t2,t3);    
    } 
    n=4*n;
         
      xmini=1e9,ymini=1e9;
      int po;
      for (int i=1;i<=n;i++){
        if (ymini>tmp[i].y||(ymini==tmp[i].y&&tmp[i].x<xmini)){
          xmini=tmp[i].x;ymini=tmp[i].y;    
          po=i;
      }
    }
    data t=tmp[po];tmp[po]=tmp[1];tmp[1]=t;
    for (int i=2;i<=n;i++) tmp[i].alp=atan2(tmp[i].y-ymini,tmp[i].x-xmini);
    a[1].x=xmini;a[1].y=ymini;
    sort(tmp+2,tmp+n+1,mycomp);
    
    cnt=1;
    tmp[1].alp=-1e9;
    for (int i=2;i<=n;i++)
      if (tmp[i].alp!=tmp[i-1].alp)
        a[++cnt]=tmp[i];                                               
    n=cnt;    
    
    int top=0;
    for (int i=1;i<=n;i++){
      while (top>=2&&X(sta[top]-sta[top-1],a[i]-sta[top-1])<-eps) 
        top--;
      sta[++top]=a[i];      
    }
    
    LDB ans=acos(-1)*2*r;
    for (int i=1;i<=top;i++)
      ans+=dis(sta[i],sta[i%top+1]);
    printf("%.2Lf
",ans);  
  }