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 ;
     return x<?-:;
 }

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

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

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