bzoj 1185 旋转卡壳 最小矩形覆盖

题目大意 就是求一个最小矩形覆盖，逆时针输出其上面的点

这里可以看出，那个最小的矩形覆盖必然有一条边经过其中凸包上的两个点，另外三条边必然至少经过其中一个点，而这样的每一个点逆时针走一遍都满足单调性

所以可以利用旋转卡壳的思想找到这样的三个点

以每一条边作为基础，循环n次得到n个这样的矩形，找到其中面积最小的即可

然后自己画画图，作出矩形对应的两条边的单位向量，那么这四个点就非常好求了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define N 50010
#define eps 1e-9
const double PI = acos(-1.0);
int n , top;

int dcmp(double x)
{
    if(fabs(x)<eps) return 0;
    return x<0?-1:1;
}

struct Point{
    double x , y;
    Point(double x=0 , double y=0):x(x),y(y){}
    void input(){scanf("%lf%lf" , &x , &y);}
    void output(){printf("%.5f %.5f
" , x , y);}
    bool operator<(const Point &m)const{
        return x<m.x||(x==m.x&&y<m.y);
    }
}po[N] , rec[N] , p[4];

typedef Point Vector;

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 b){return Vector(a.x*b , a.y*b);}
Vector operator/(Vector a , double b){return Vector(a.x/b , a.y/b);}
double operator*(Vector a , Vector b){return a.x*b.y-a.y*b.x;}

double Dot(Vector a , Vector b){return a.x*b.x+a.y*b.y;}
double Len(Vector a){return sqrt(Dot(a,a));}
int Polxy()
{
    sort(po , po+n);
    rec[0]=po[0] , rec[1]=po[1];
    top=2;
    for(int i=2 ; i<n ; i++){
        while(top>=2&&(rec[top-1]-rec[top-2])*(po[i]-rec[top-2])<=0)
            top--;
        rec[top++] = po[i];
    }
    int tmp=top;
    for(int i=n-2 ; i>=0 ; i--){
        while(top>=tmp&&(rec[top-1]-rec[top-2])*(po[i]-rec[top-2])<=0)
            top--;
        rec[top++] = po[i];
    }
    top--;
    return top;
}

Vector Normal(Vector a)
{
    double l = Len(a);
    return Vector(-a.y/l , a.x/l);
}

double calCalip()
{
  //  for(int i=0 ; i<top ; i++) rec[i].output();
    Point ch[2];
    ch[0] = rec[0] , ch[1] = rec[1];
    int i1=0 , i2=0 , i3=0;
    double maxn = 1e18;
    ch[0] = rec[0] , ch[1] = rec[1];
    while(dcmp(fabs((rec[(i1+1)%top]-ch[0])*(ch[1]-ch[0]))-fabs((rec[i1]-ch[0])*(ch[1]-ch[0])))>=0) i1=(i1+1)%top;
    while(dcmp(Dot(rec[(i2+1)%top]-ch[0] , ch[1]-ch[0])-Dot(rec[i2]-ch[0] , ch[1]-ch[0]))>=0) i2=(i2+1)%top;
    while(dcmp(Dot(rec[(i3-1+top)%top]-ch[1] , ch[0]-ch[1])-Dot(rec[i3]-ch[1] , ch[0]-ch[1]))>=0) i3=(i3-1+top)%top;
    for(int i=0 ; i<top ; i++){
        ch[0] = rec[i] , ch[1] = rec[(i+1)%top];
        while(dcmp(fabs((rec[(i1+1)%top]-ch[0])*(ch[1]-ch[0]))-fabs((rec[i1]-ch[0])*(ch[1]-ch[0])))>=0) i1=(i1+1)%top;
        while(dcmp(Dot(rec[(i2+1)%top]-ch[0] , ch[1]-ch[0])-Dot(rec[i2]-ch[0] , ch[1]-ch[0]))>=0) i2=(i2+1)%top;
        while(dcmp(Dot(rec[(i3+1)%top]-ch[1] , ch[0]-ch[1])-Dot(rec[i3]-ch[1] , ch[0]-ch[1]))>=0) i3=(i3+1)%top;
        double l = Len(ch[1]-ch[0]);
        double h = fabs((rec[i1]-ch[0])*(ch[1]-ch[0]))/l;
        double len1 = Dot(rec[i2]-ch[1] , ch[1]-ch[0])/l; //右侧长度
        double len2 = Dot(rec[i3]-ch[0] , ch[0]-ch[1])/l; //左侧长度
        double suml = l+len1+len2;
       // cout<<i<<" "<<l<<" "<<h<<" "<<len1<<" "<<len2<<" "<<suml*h<<endl;
        if(dcmp(suml*h-maxn)<0){
            Vector unit1 = (ch[1]-ch[0])/l;
            Vector unit2 = Normal(unit1);
            maxn = suml*h;
            p[0] = ch[1]+unit1*len1;
            p[1] = p[0]+unit2*h;
            p[2] = p[1]-unit1*suml;
            p[3] = p[2]-unit2*h;
        }
    }
    return maxn;
}

int main()
{
   // freopen("in.txt" , "r" , stdin);
    while(scanf("%d" , &n)!=EOF)
    {
        for(int i=0 ; i<n ; i++) po[i].input();
        Polxy();
        double ret = calCalip();
        printf("%.5f
" , ret);
        int st = 0;
        for(int i=1 ; i<4 ; i++){
            if(p[i].y<p[st].y || (p[i].y==p[st].y&&p[i].x<p[st].x)) st = i;
        }
        for(int i=0;i<4;i++){
            p[st].output();
            st = (st+1)%4;
        }
    }
    return 0;
}