UVA 10173 Smallest Bounding Rectangle（最小外接矩形）

题目链接：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1114

题意：求多边形最小外接矩形。

思路：模板。

#include <iostream>

#include <cstdio>

#include <cmath>

#include <algorithm>

#define min(x,y) ((x)<(y)?(x):(y))

using namespace std;





struct point

{

    double x,y;



    point(){}

    point(double _x,double _y)

    {

        x=_x;

        y=_y;

    }





    void get()

    {

        scanf("%lf%lf",&x,&y);

    }

};



const double EPS=1e-8;

const int MAX=1005;

point p[MAX],q[MAX],temp;

int n,m;





int DB(double x)

{

    if(x>EPS) return 1;

    if(x<-EPS) return -1;

    return 0;

}







double Dis(point a,point b)

{

    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));

}



//判断p在向量ab的哪一侧

//右侧：返回正值

//左侧：返回负值

//在向量ab上返回0

double cross(point a,point b,point p)

{

    return (b.x-a.x)*(p.y-a.y)-(b.y-a.y)*(p.x-a.x);

}





int cmp(point a,point b)

{

    double x=Dis(a,temp),y=Dis(b,temp);

    int flag=DB(cross(temp,a,b));

    if(flag) return flag==1;

    return DB(x-y)<=0;

}



void Graham(point p[],int n,point q[],int &m)

{

    point t;

    int i,k=0,a,b;

    for(i=1;i<n;i++)

    {

        a=DB(p[i].y-p[k].y);

        b=DB(p[i].x-p[k].x);

        if(a==-1||!a&&b==-1) k=i;

    }

    if(k!=0) t=p[0],p[0]=p[k],p[k]=t;

	temp=p[0];

    sort(p+1,p+n,cmp);

    q[0]=p[0];

    q[1]=p[1];

    p[n]=p[0];

    m=2;

    for(i=2;i<=n;i++)

    {

        while(m>1&&DB(cross(q[m-2],q[m-1],p[i]))<=0) m--;

        q[m++]=p[i];

    }

    m--;

}







//向量点乘

//ab*ac

double dot(point a,point b,point c)

{

    return (c.x-a.x)*(b.x-a.x)+(c.y-a.y)*(b.y-a.y);

}



double calMinRect(point pt[],int n)

{

    if(n<3) return 0;

    int i,p=1,q=1,r;

    double ans=1e10,a,b,c;

    for(i=0;i<n;i++)

    {

        while(1)

        {

            a=cross(pt[i],pt[i+1],pt[p+1]);

            b=cross(pt[i],pt[i+1],pt[p]);

            if(DB(a-b)<=0) break;

            p=(p+1)%n;

        }

        while(1)

        {

            a=dot(pt[i],pt[i+1],pt[q+1]);

            b=dot(pt[i],pt[i+1],pt[q]);

            if(DB(a-b)<=0) break;

            q=(q+1)%n;

        }

        if(!i)r=q;

        while(1)

        {

            a=dot(pt[i],pt[i+1],pt[r+1]);

            b=dot(pt[i],pt[i+1],pt[r]);

            if(DB(a-b)>0) break;

            r=(r+1)%n;

        }

        a=cross(pt[i],pt[i+1],pt[p]);

        b=dot(pt[i],pt[i+1],pt[q])-dot(pt[i],pt[i+1],pt[r]);

        c=dot(pt[i],pt[i+1],pt[i+1]);

        ans=min(ans,a*b/c);

    }

    return ans;

}



int main()

{

	while(scanf("%d",&n),n)

	{

		int i;

		for(i=0;i<n;i++) p[i].get();

		Graham(p,n,q,m);

	    double ans=calMinRect(q,m);

		printf("%.4lf\n",ans);

	}

	return 0;

}