BZOJ1185 : [HNOI2007]最小矩形覆盖

求出凸包后，矩形的一条边一定与凸包的某条边重合。

枚举每条边，求出离它最远的点和离它最左最右的点，因为那三个点是单调变化的，所以复杂度为$O(n)$。

注意精度。

 

#include<cstdio>
#include<algorithm>
#include<cmath>
#define N 50010
using namespace std;
typedef double D;
struct P{D x,y;P(){}P(D _x,D _y){x=_x,y=_y;}}p[N],pp[N],hull[N],pivot,A,B,C,rect[8];
int n,i,j,l,r,k;
D w,h,ans=1e20,tmp,len;
bool del[N];
inline int zero(D x){return fabs(x)<1e-4;}
inline int sig(D x){if(fabs(x)<1e-8)return 0;return x>0?1:-1;}
inline D cross(P A,P B,P C){return(B.x-A.x)*(C.y-A.y)-(B.y-A.y)*(C.x-A.x);}
inline D distsqr(P A,P B){return(A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y);}
inline bool cmp(P a,P b){
  D t=cross(pivot,a,b);
  return sig(t)==1||sig(t)==0&&sig(distsqr(pivot,a)-distsqr(pivot,b))==-1;
}
inline void convexhull(int n,P stck[],int&m){
  int i,k,top;
  for(i=0;i<n;i++)pp[i]=p[i];
  for(k=0,i=1;i<n;i++)if(pp[i].y<pp[k].y||(pp[i].y==pp[k].y&&pp[i].x<pp[k].x))k=i;
  pivot=pp[k];pp[k]=pp[0];pp[0]=pivot;
  sort(pp+1,pp+n,cmp);
  stck[0]=pp[0];stck[1]=pp[1];
  for(top=1,i=2;i<n;i++){
    while(top&&sig(cross(pp[i],stck[top],stck[top-1]))>=0)--top;
    stck[++top]=pp[i];
  }
  m=top+1;
}
inline D area(P A,P B,P C){return fabs(cross(A,B,C));}
inline P vertical(P A,P B){return P(A.x-B.y+A.y,A.y+B.x-A.x);}
int main(){
  scanf("%d",&n);
  for(i=0;i<n;i++)scanf("%lf%lf",&p[i].x,&p[i].y);
  convexhull(n,hull,n);
  for(i=1;i<n;i++)if(zero(hull[i].x-hull[i-1].x)&&zero(hull[i].y-hull[i-1].y))del[i]=1;
  for(k=i=0;i<n;i++)if(!del[i])hull[k++]=hull[i];
  for(hull[n=k]=hull[i=0];i<n;i++){
    A=hull[i],B=hull[i+1],C=vertical(A,B);
    while(sig(area(A,B,hull[j])-area(A,B,hull[j+1]))<1)j=(j+1)%n;
    while(sig(cross(A,C,hull[l])-cross(A,C,hull[l+1]))<1)l=(l+1)%n;
    while(sig(cross(A,C,hull[r])-cross(A,C,hull[r+1]))>-1)r=(r+1)%n;
    len=sqrt(distsqr(A,B));
    h=area(A,B,hull[j])/len;
    w=(cross(A,C,hull[l])-cross(A,C,hull[r]))/len;
    if(sig(h*w-ans)==-1){
      ans=h*w;
      tmp=area(A,B,hull[l])/len/len;
      rect[0]=P(hull[l].x+tmp*(A.x-C.x),hull[l].y+tmp*(A.y-C.y));
      tmp=h/len;
      rect[3]=P(rect[0].x+tmp*(C.x-A.x),rect[0].y+tmp*(C.y-A.y));
      tmp=w/len;
      rect[1]=P(rect[0].x+tmp*(B.x-A.x),rect[0].y+tmp*(B.y-A.y));
      rect[2]=P(rect[3].x+tmp*(B.x-A.x),rect[3].y+tmp*(B.y-A.y));
    }
  }
  for(i=0;i<4;i++)rect[i+4]=rect[i];
  for(j=0,i=1;i<4;i++)if(sig(rect[i].y-rect[j].y)==-1||sig(rect[i].y-rect[j].y)==0&&sig(rect[i].x-rect[j].x)==-1)j=i;
  printf("%.0f.00000\n",ans);
  for(i=0;i<4;i++)printf("%.0f.00000 %.0f.00000\n",rect[j+i].x,rect[j+i].y);
  return 0;
}