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;

}