【HNOI2007】bzoj1185 最小矩形覆盖

来源：互联网发布：xp系统网络连接错误678 编辑：程序博客网时间：2024/06/08 10:06

首先可以发现，矩形的一条边一定在凸包上。于是只需要旋转着枚举这条边，同时维护矩形另外三条边夹住的点。可以发现这三个点的移动都是单调的，于是复杂度O(n)。

#include<cstdio>#include<cmath>#include<algorithm>using namespace std;const double eps=1e-8,oo=1e16;const int maxn=50010;int cmp(double x){    if (x>eps) return 1;    if (fabs(x)<=eps) return 0;    return -1;}struct Vector{    double x,y;    void rd()    {        scanf("%lf%lf",&x,&y);    }    void prt()    {        printf("%.5f %.5f\n",x,y);    }    bool operator < (const Vector &v) const    {        int tem=cmp(x-v.x);        if (tem) return tem==-1;        return cmp(y-v.y)==-1;    }    Vector operator + (const Vector &v) const    {        return (Vector){x+v.x,y+v.y};    }    Vector operator - (const Vector &v) const    {        return (Vector){x-v.x,y-v.y};    }    Vector operator * (const double &k) const    {        return (Vector){x*k,y*k};    }}a[maxn],f[maxn],res[4],now[4],v1,v2;typedef Vector Point;double dot(Vector v,Vector u){    return v.x*u.x+v.y*u.y;}double cross(Vector v,Vector u){    return v.x*u.y-v.y*u.x;}Vector normal(Vector v){    return (Vector){v.y,-v.x};}double len(Vector v){    return sqrt(dot(v,v));}int cmpy(int x,int y){    int tem=cmp(res[x].y-res[y].y);    if (tem) return tem==-1;    return cmp(res[x].x-res[y].x)==-1;}struct Line{    Point p;    Vector v;    Point get(double k)    {        return p+v*k;    }};Point intersection(Line l1,Line l2){    Vector w=l1.p-l2.p;    return l1.get(cross(l2.v,w)/cross(l1.v,l2.v));}double ans=oo;int n,m;int main(){    int mm,u;    scanf("%d",&n);    for (int i=1;i<=n;i++) a[i].rd();    sort(a+1,a+n+1);    f[0]=a[1];    m=1;    for (int i=2;i<=n;i++)    {        while (m>1&&cmp(cross(f[m-1]-f[m-2],a[i]-f[m-2]))>=0) m--;        f[m++]=a[i];    }    mm=m;    for (int i=n-1;i;i--)    {        while (m>mm&&cmp(cross(f[m-1]-f[m-2],a[i]-f[m-2]))>=0) m--;        f[m++]=a[i];    }    m--;    for (int i=0,j=1,x=1,y=-1;i<m;i++)    {        while (cmp(cross(f[i+1]-f[i],f[(j+1)%m]-f[j]))<=0) j=(j+1)%m;        while (cmp(dot(f[i+1]-f[i],f[(x+1)%m]-f[x]))>=0) x=(x+1)%m;        if (y==-1) y=j;        while (cmp(dot(f[i+1]-f[i],f[(y+1)%m]-f[y]))<=0) y=(y+1)%m;        v1=f[i+1]-f[i];        v2=normal(v1);        now[0]=intersection((Line){f[j],v1},(Line){f[x],v2});        now[1]=intersection((Line){f[i],v1},(Line){f[x],v2});        now[2]=intersection((Line){f[i],v1},(Line){f[y],v2});        now[3]=intersection((Line){f[j],v1},(Line){f[y],v2});        if (cmp(len(now[0]-now[1])*len(now[1]-now[2])-ans)==-1)        {            ans=len(now[0]-now[1])*len(now[1]-now[2]);            for (int k=0;k<4;k++) res[k]=now[k];        }    }    printf("%.5f\n",ans);    u=0;    for (int i=1;i<4;i++) u=min(u,i,cmpy);    for (int i=u,cnt=0;cnt<=3;i=(i+1)%4,cnt++) res[i].prt();}

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