BZOJ 1185: [HNOI2007]最小矩形覆盖 [旋转卡壳]

竟然1A了........哈哈哈哈哈哈哈哈哈哈

首先猜有一条边是凸边上的边(理由：不是的话我不会做)

然后旋转卡壳，最上面就是距离最远的点，最右面是点积最大，最左面是点积最小

然后就是各种向量运算我用了v和u分别是宽和高方向的单位向量，感觉挺方便的....

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;
typedef long long ll;
const int N=5e4+5,INF=1e9;
const double eps=1e-8;

inline int sgn(double x){
    if(abs(x)<eps) return 0;
    else return x<0?-1:1;
}

struct Vector{
    double x,y;
    Vector(double a=0,double b=0):x(a),y(b){}
    bool operator <(const Vector &a)const{
        return sgn(y-a.y)<0||(sgn(y-a.y)==0&&sgn(x-a.x)<0);
    }
};
typedef Vector Point;
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);}
bool operator ==(Vector a,Vector b){return sgn(a.x-b.x)==0&&sgn(a.y-b.y)==0;}
double Dot(Vector a,Vector b){return a.x*b.x+a.y*b.y;}
double Cross(Vector a,Vector b){return a.x*b.y-a.y*b.x;}

double Len(Vector a){return sqrt(Dot(a,a));}
double Len2(Vector a){return Dot(a,a);}
double DisTL(Point p,Point a,Point b){
    Vector v1=p-a,v2=b-a;
    return abs(Cross(v1,v2)/Len(v2));
}
int ConvexHull(Point p[],int n,Point ch[]){
    sort(p+1,p+1+n);
    int m=0;
    for(int i=1;i<=n;i++){
        while(m>1&&sgn(Cross(ch[m]-ch[m-1],p[i]-ch[m-1]))<=0) m--;
        ch[++m]=p[i];
    }
    int k=m;
    for(int i=n-1;i>=1;i--){
        while(m>k&&sgn(Cross(ch[m]-ch[m-1],p[i]-ch[m-1]))<=0) m--;
        ch[++m]=p[i];
    }
    if(n>1) m--;
    return m;
}
Point pos[5];
void RotatingCalipers(Point p[],int n){
    double ans=INF;
    p[n+1]=p[1];
    int j=1,k=1,l=1;
    for(int i=1;i<=n;i++){
        while(sgn(DisTL(p[j],p[i],p[i+1])-DisTL(p[j+1],p[i],p[i+1]))<=0) j=j%n+1;
        while(sgn(Dot(p[k]-p[i],p[i+1]-p[i])-Dot(p[k+1]-p[i],p[i+1]-p[i]))<=0) k=k%n+1;
        if(i==1) l=j;
        while(sgn(Dot(p[l]-p[i],p[i+1]-p[i])-Dot(p[l+1]-p[i],p[i+1]-p[i]))>=0) l=l%n+1;
        double len=Len(p[i+1]-p[i]);
        double h=DisTL(p[j],p[i],p[i+1]),
                w=Dot(p[k]-p[i],p[i+1]-p[i])/len+abs(Dot(p[l]-p[i],p[i+1]-p[i])/len);
        if(h*w<ans){
            Vector v=(p[i+1]-p[i])/len;
            Vector u(-v.y,v.x);
            ans=h*w;
            pos[1]=v*Dot(p[l]-p[i],v)+p[i];
            pos[2]=v*Dot(p[k]-p[i],v)+p[i];
            pos[3]=pos[2]+u*h;
            pos[4]=pos[1]+u*h;
        }
    }
    int mn=1;
    for(int i=2;i<=4;i++) if(pos[i]<pos[mn]) mn=i;
    printf("%lf
",ans);
    for(int i=mn;i<=4;i++) printf("%lf %lf
",pos[i].x,pos[i].y);
    for(int i=1;i<mn;i++) printf("%lf %lf
",pos[i].x,pos[i].y);
}

int n;
Point p[N],ch[N];
int main(int argc, const char * argv[]) {
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%lf%lf",&p[i].x,&p[i].y);
    n=ConvexHull(p,n,ch);
    RotatingCalipers(ch,n);
    
}

BZOJ 1185: [HNOI2007]最小矩形覆盖 [旋转卡壳]

1185: [HNOI2007]最小矩形覆盖

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