1185: [HNOI2007]最小矩形覆盖

题目描述

给定一些点的坐标，要求求能够覆盖所有点的最小面积的矩形，

输出所求矩形的面积和四个顶点坐标

 

输入

第一行为一个整数n（3<=n<=50000）

从第2至第n+1行每行有两个浮点数，表示一个顶点的x和y坐标，不用科学计数法

 

输出

第一行为一个浮点数，表示所求矩形的面积（精确到小数点后5位），

接下来4行每行表示一个顶点坐标，要求第一行为y坐标最小的顶点，

其后按逆时针输出顶点坐标.如果用相同y坐标，先输出最小x坐标的顶点

样例输入

6 1.0 3.00000
1 4.00000
2.0000 1
3 0.0000
3.00000 6
6.0 3.0

样例输出

18.00000
3.00000 0.00000
6.00000 3.00000
3.00000 6.00000
0.00000 3.00000

 

 

 

 

暴力过的，先把坐标旋转，计算新坐标系下所有点的新的坐标，找到4个角，然后把4个角的坐标转成原坐标系下的坐标，时间复杂度O(n^2)，涉及角度转换，时间常数也很大

#include<iostream>
#include<stdio.h>
#include<math.h>
#include<algorithm>
using namespace std;

typedef struct {
    double x,y;
}Point;
Point p[50005];
int n;
int Q[50005],top;
int ans[50005];

bool cmp(Point a,Point b){
    return a.y<b.y||(a.y==b.y&&a.x<b.x);
}

double cross(int i,int j,int k){
    Point a=p[i];
    Point b=p[j];
    Point c=p[k];
    return (c.x-b.x)*(a.y-b.y)-(c.y-b.y)*(a.x-b.x);

}

void scan(){
    sort(p+1,p+1+n,cmp);
    
    Q[1]=1;Q[2]=2;top=2;
    
    for(int i=3;i<=n;++i)
    {
        while(top>1&&cross(Q[top-1],Q[top],i)<=0) top--;
        Q[++top]=i;
    }
    
    ans[0]=top;
    for(int i=1;i<=top;++i)
    ans[i]=Q[i];
    
    Q[1]=n;Q[2]=n-1;top=2;
    for(int i=n-2;i>=1;--i)
    {
        while(top>1&&cross(Q[top-1],Q[top],i)<=0) top--;
        Q[++top]=i;
    }
    
    for(int i=2;i<top;++i)
    {
        ans[0]++;
        ans[ans[0]]=Q[i];
    }

}

double computeTheta(int i,int j){
    Point a=p[i];
    Point b=p[j];
    if(a.y>b.y) {Point c=a;a=b;b=c;}
//    cout<<(b.x-a.x)<<" "<<sqrt((b.x-a.x)*(b.x-a.x)+(b.y-a.y)*(b.y-a.y))<<"  "<<(b.x-a.x)/sqrt((b.x-a.x)*(b.x-a.x)+(b.y-a.y)*(b.y-a.y))<<"  "<<acos((b.x-a.x)/sqrt((b.x-a.x)*(b.x-a.x)+(b.y-a.y)*(b.y-a.y)))<<endl;
    return acos((b.x-a.x)/sqrt((b.x-a.x)*(b.x-a.x)+(b.y-a.y)*(b.y-a.y))); 
}
bool isBetter(Point tempPoint[],Point ansPoint[]){
    for(int i=0;i<3;++i)
    {
        if(tempPoint[i].y<ansPoint[i].y||(tempPoint[i].y==ansPoint[i].y&&tempPoint[i].x<ansPoint[i].x)) return true;
        if(tempPoint[i].y>ansPoint[i].y||(tempPoint[i].y==ansPoint[i].y&&tempPoint[i].x>ansPoint[i].x)) return false;
    }
    
    return false;

}
int main()
{
    scanf("%d",&n);
    
    for(int i=1;i<=n;++i)
    scanf("%lf %lf",&p[i].x,&p[i].y);

    scan(); 
    
    ans[0]++;
    ans[ans[0]]=1;
    
    double x,y; 
    double minarea;
    bool initialArea=true;
    double minx,maxx,miny,maxy;
    bool initialXY=true;
    Point ansPoint[4];
    
    for(int i=1;i<ans[0];++i)
    {
    //    cout<<p[ans[i]].x<<" "<<p[ans[i]].y<<" "<<p[ans[i+1]].x<<" "<<p[ans[i+1]].y<<endl;
        double theta=computeTheta(ans[i],ans[i+1]);
        initialXY=true;
        for(int j=1;j<ans[0];++j)
        {
            x=p[ans[j]].x*cos(theta)+p[ans[j]].y*sin(theta);
            y=p[ans[j]].y*cos(theta)-p[ans[j]].x*sin(theta);
            if(initialXY==true)
            {
                minx=x;maxx=x;miny=y;maxy=y;
                initialXY=false;
            }
            if(initialArea==false&&(maxx-minx)*(maxy-miny)>minarea+0.5) break;
            if(x<minx) minx=x;
            if(x>maxx) maxx=x;
            if(y<miny) miny=y;
            if(y>maxy) maxy=y;
        }
        
    //    cout<<theta<<"  "<<(maxx-minx)*(maxy-miny)<<"  fdg"<<endl;
        
        if(initialArea==true||(maxx-minx)*(maxy-miny)<minarea)
        {
            initialArea=false;
            minarea=(maxx-minx)*(maxy-miny);
            ansPoint[0].x=minx*cos(theta)-miny*sin(theta);
            ansPoint[0].y=minx*sin(theta)+miny*cos(theta);
            
            ansPoint[1].x=minx*cos(theta)-maxy*sin(theta);
            ansPoint[1].y=minx*sin(theta)+maxy*cos(theta);
            
            ansPoint[2].x=maxx*cos(theta)-miny*sin(theta);
            ansPoint[2].y=maxx*sin(theta)+miny*cos(theta);
            
            ansPoint[3].x=maxx*cos(theta)-maxy*sin(theta);
            ansPoint[3].y=maxx*sin(theta)+maxy*cos(theta);
        }
        else if(initialArea==false&&fabs((maxx-minx)*(maxy-miny)-minarea)<1e-6)
        {
        //    cout<<"NOW  "<<(maxx-minx)*(maxy-miny)<<endl;
            Point tempPoint[4];
            tempPoint[0].x=minx*cos(theta)-miny*sin(theta);
            tempPoint[0].y=minx*sin(theta)+miny*cos(theta);
            
            tempPoint[1].x=minx*cos(theta)-maxy*sin(theta);
            tempPoint[1].y=minx*sin(theta)+maxy*cos(theta);
            
            tempPoint[2].x=maxx*cos(theta)-miny*sin(theta);
            tempPoint[2].y=maxx*sin(theta)+miny*cos(theta);
            
            tempPoint[3].x=maxx*cos(theta)-maxy*sin(theta);
            tempPoint[3].y=maxx*sin(theta)+maxy*cos(theta);
            
            sort(ansPoint,ansPoint+4,cmp);
            sort(tempPoint,tempPoint+4,cmp);
            if(isBetter(tempPoint,ansPoint))
            {
                ansPoint[0]=tempPoint[0];
                ansPoint[1]=tempPoint[1];
                ansPoint[2]=tempPoint[2];
                ansPoint[3]=tempPoint[3];
            }
            
            
        
        }
        
    }
    
    printf("%.5f
",minarea);
    for(int i=0;i<=3;++i)
    {
        if(fabs(ansPoint[i].x)<1e-6)
        ansPoint[i].x=0;
        if(fabs(ansPoint[i].y)<1e-6)
        ansPoint[i].y=0;
    }
    
    n=4;
    for(int i=1;i<=4;++i)
    p[i]=ansPoint[i-1];
    scan();
    
    for(int i=1;i<=4;++i)
    printf("%.5f %.5f
",p[ans[i]].x,p[ans[i]].y);

    return 0;

} 

看到其他人的题解，挺好的，旋转卡壳，时间复杂度O(N)，下面这个代码很优雅，侵删

#include<bits/stdc++.h>
#define db double
using namespace std;
const int M=5e4+5;
const db eps=1e-8;
struct pt{db x,y;};
int sig(db x){return (x>eps)-(x<-eps);}
pt operator -(pt a,pt b){return (pt){a.x-b.x,a.y-b.y};}
pt operator +(pt a,pt b){return (pt){a.x+b.x,a.y+b.y};}
db operator *(pt a,pt b){return a.x*b.y-a.y*b.x;}
db operator /(pt a,pt b){return a.x*b.x+a.y*b.y;}
pt operator *(pt a,db b){return (pt){a.x*b,a.y*b};}
pt operator /(pt a,db b){return (pt){a.x/b,a.y/b};}
bool operator <(pt a,pt b){return !sig(a.y-b.y)?a.x<b.x:a.y<b.y;}
db area(pt a,pt b,pt c){return (b-a)*(c-a);}
db shad(pt a,pt b,pt c){return (b-a)/(c-a);}
db sqr(db x){return x*x;}
db dis(pt a,pt b){return sqrt(sqr(a.x-b.x)+sqr(a.y-b.y));}
bool cmp(pt a,pt b){return sig(a.x-b.x)?a.y<b.y:a.x<b.x;}
pt p[M],sta[M],rec[5];
int top,n;
void in()
{
    scanf("%d",&n);
    for(int i=1;i<=n;++i)
    scanf("%lf%lf",&p[i].x,&p[i].y);
}
void tubao()
{
    top=-1;
    sort(p+1,p+1+n);
    for(int i=1;i<=n;++i)
    {
        while(top>0&&sig(area(sta[top-1],sta[top],p[i]))<=0)--top;
        sta[++top]=p[i];
    }
    int k=top;
    for(int i=n-1;i>=1;--i)
    {
        while(top>k&&sig(area(sta[top-1],sta[top],p[i]))<=0)--top;
        sta[++top]=p[i];
    }
}
int x[M];
db fuck(db x){return !sig(x)?fabs(x):x;}
void ac()
{
    int le=1,ri=1,up=1;
    db L,R,H,D,tmp,ans=1e60;
    for(int i=0;i<top;++i)
    {
        D=dis(sta[i],sta[i+1]);
        while(sig(area(sta[i],sta[i+1],sta[up])-area(sta[i],sta[i+1],sta[up+1]))<=0)up=(up+1)%top;
        while(sig(shad(sta[i],sta[i+1],sta[ri])-shad(sta[i],sta[i+1],sta[ri+1]))<=0)ri=(ri+1)%top;
        if(i==0)le=up;
        while(sig(shad(sta[i],sta[i+1],sta[le])-shad(sta[i],sta[i+1],sta[le+1]))>=0)le=(le+1)%top;
        L=shad(sta[i],sta[i+1],sta[le])/D;
        R=shad(sta[i],sta[i+1],sta[ri])/D;
        H=area(sta[i],sta[i+1],sta[up])/D;
        H=H>0?H:-H;
        tmp=(R-L)*H;
        if(tmp<ans)
        {
            ans=tmp;
            rec[0]=sta[i]+(sta[i+1]-sta[i])*(R/D);
            rec[1]=rec[0]+(sta[ri]-rec[0])*(H/dis(sta[ri],rec[0]));
            rec[2]=rec[1]-(rec[0]-sta[i])*((R-L)/dis(sta[i],rec[0]));
            rec[3]=rec[2]-(rec[1]-rec[0]);
        }
    }
    ans=fabs(ans);
    printf("%.5lf
",ans);
    int fir=0;
    for(int i=1;i<=3;++i)
    if(rec[i]<rec[fir])fir=i;
    for(int i=0;i<=3;++i)
    printf("%.5lf %.5lf
",fuck(rec[(i+fir)%4].x),fuck(rec[(i+fir)%4].y));
}
int main()
{
    in();tubao();
    ac();
    return 0;
}