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+5;
const double eps=1e-8;
double ans=1e60;
struct Vector{
    double x,y;
    Vector(double x=0,double y=0):x(x),y(y){}
}p[N],q[N],t[5];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[1])*(b-p[1]);
    if(fabs(t)<eps) return Length(p[1]-a)-Length(p[1]-b)<0;
    else return t>0;
}
void graham(){//凸包 
    for(int i=2;i<=n;i++) if(p[i]<p[1]) swap(p[i],p[1]);
    sort(p+2,p+n+1,cmp);
    q[++top]=p[1];
    for(int i=2;i<=n;i++){
        while(top>1&&(q[top]-q[top-1])*(p[i]-q[top])<eps) top--;
        q[++top]=p[i];
    }
    q[0]=q[top];
}
void RC(){//旋转卡壳 
    int l=1,r=1,p=1;
    double L,R,D,H;
    for(int i=0;i<top;i++){
        D=Length(q[i]-q[i+1]);
        while((q[i+1]-q[i])*(q[p+1]-q[i])-(q[i+1]-q[i])*(q[p]-q[i])>-eps) p=(p+1)%top;
        while((q[i+1]-q[i])/(q[r+1]-q[i])-(q[i+1]-q[i])/(q[r]-q[i])>-eps) r=(r+1)%top;
        if(i==0) l=r;
        while((q[i+1]-q[i])/(q[l+1]-q[i])-(q[i+1]-q[i])/(q[l]-q[i])<eps) l=(l+1)%top;
            L=(q[i+1]-q[i])/(q[l]-q[i])/D,R=(q[i+1]-q[i])/(q[r]-q[i])/D;
        H=(q[i+1]-q[i])*(q[p]-q[i])/D;
        if(H<0) H=-H;
        double tmp=(R-L)*H;
        if(tmp<ans){
            ans=tmp;
            t[0]=q[i]+(q[i+1]-q[i])*(R/D);
            t[1]=t[0]+(q[r]-t[0])*(H/Length(t[0]-q[r]));
            t[2]=t[1]-(t[0]-q[i])*((R-L)/Length(q[i]-t[0]));
            t[3]=t[2]-(t[1]-t[0]);
        }
    }
}
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%lf%lf",&p[i].x,&p[i].y);
    graham();
    RC();
    printf("%.5lf\n",ans);
    int fir=0;
    for(int i=1;i<4;i++) if(t[i]<t[fir]) fir=i;
    for(int i=0;i<4;i++){//某些OJ评测卡精度.. 比如，洛谷
        if(fabs(t[(i+fir)%4].x)<1e-6) printf("0.00000 ");else printf("%.5lf ",t[(i+fir)%4].x);
        if(fabs(t[(i+fir)%4].y)<1e-6) printf("0.00000\n");else printf("%.5lf\n",t[(i+fir)%4].y);
    }
    return 0;
}
/*
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
*/

__EOF__

本文作者：shenben
本文链接：https://www.cnblogs.com/shenben/p/6285592.html
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