1185: [HNOI2007]最小矩形覆盖

1185: [HNOI2007]最小矩形覆盖

Time Limit: 10 Sec  Memory Limit: 162 MBSec  Special Judge
Submit: 1426  Solved: 648
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Description

Input

 

Output

 

Sample Input

 

Sample Output

 

HINT

 

Source

计算几何 vfleaking提供Spj

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const int N=1e5+;
const double eps=1e-;
double ans=1e60;
struct Vector{
    double x,y;
    Vector(double x=,double y=):x(x),y(y){}
}p[N],q[N],t[];int n,top;
Vector operator + (Vector A,Vector B){return Vector(A.x+B.x,A.y+B.y);}
Vector operator - (Vector A,Vector B){return Vector(A.x-B.x,A.y-B.y);}
Vector operator * (Vector A,double p){return Vector(A.x*p,A.y*p);}
Vector operator / (Vector A,double p){return Vector(A.x/p,A.y/p);}
double operator * (Vector A,Vector B){return A.x*B.y-A.y*B.x;;}
double operator / (Vector A,Vector B){return A.x*B.x+A.y*B.y;;}
bool operator<(Vector a,Vector b){
    return fabs(a.y-b.y)<eps?a.x<b.x:a.y<b.y;
}
bool operator==(Vector a,Vector b){
    return fabs(a.x-b.x)<eps&&fabs(a.y-b.y)<eps;
}
double Dot(Vector A,Vector B){return A.x*B.x+A.y*B.y;}
double Length(Vector A){return sqrt(Dot(A,A));}
double Angle(Vector A,Vector B){return acos(Dot(A,B)/(Length(A)*Length(B)));}
double Cross(Vector A,Vector B){return A.x*B.y-A.y*B.x;}
double Area2(Vector A,Vector B,Vector C){return Cross(B-A,C-A);}
bool cmp(Vector a,Vector b){
    double t=(a-p[])*(b-p[]);
    if(fabs(t)<eps) return Length(p[]-a)-Length(p[]-b)<;
    else return t>;
}
void graham(){//凸包
    for(int i=;i<=n;i++) if(p[i]<p[]) swap(p[i],p[]);
    sort(p+,p+n+,cmp);
    q[++top]=p[];
    for(int i=;i<=n;i++){
        while(top>&&(q[top]-q[top-])*(p[i]-q[top])<eps) top--;
        q[++top]=p[i];
    }
    q[]=q[top];
}
void RC(){//旋转卡壳
    int l=,r=,p=;
    double L,R,D,H;
    for(int i=;i<top;i++){
        D=Length(q[i]-q[i+]);
        while((q[i+]-q[i])*(q[p+]-q[i])-(q[i+]-q[i])*(q[p]-q[i])>-eps) p=(p+)%top;
        while((q[i+]-q[i])/(q[r+]-q[i])-(q[i+]-q[i])/(q[r]-q[i])>-eps) r=(r+)%top;
        if(i==) l=r;
        while((q[i+]-q[i])/(q[l+]-q[i])-(q[i+]-q[i])/(q[l]-q[i])<eps) l=(l+)%top;
            L=(q[i+]-q[i])/(q[l]-q[i])/D,R=(q[i+]-q[i])/(q[r]-q[i])/D;
        H=(q[i+]-q[i])*(q[p]-q[i])/D;
        if(H<) H=-H;
        double tmp=(R-L)*H;
        if(tmp<ans){
            ans=tmp;
            t[]=q[i]+(q[i+]-q[i])*(R/D);
            t[]=t[]+(q[r]-t[])*(H/Length(t[]-q[r]));
            t[]=t[]-(t[]-q[i])*((R-L)/Length(q[i]-t[]));
            t[]=t[]-(t[]-t[]);
        }
    }
}
int main(){
    scanf("%d",&n);
    for(int i=;i<=n;i++) scanf("%lf%lf",&p[i].x,&p[i].y);
    graham();
    RC();
    printf("%.5lf\n",ans);
    int fir=;
    for(int i=;i<;i++) if(t[i]<t[fir]) fir=i;
    for(int i=;i<;i++){//某些OJ评测卡精度.. 比如，洛谷
        if(fabs(t[(i+fir)%].x)<1e-) printf("0.00000 ");else printf("%.5lf ",t[(i+fir)%].x);
        if(fabs(t[(i+fir)%].y)<1e-) printf("0.00000\n");else printf("%.5lf\n",t[(i+fir)%].y);
    }
    return ;
}
/*
Sample Input
6
1.0 3.00000
1 4.00000
2.00000 1
3 0.00000
3.00000 6
6.0 3.0

Sample Output
18.00000
3.00000 0.00000
6.00000 3.00000
3.00000 6.00000
0.00000 3.00000
*/