UVA10173(求凸包的面积最小外接矩形)

题目：10173 - Smallest Bounding Rectangle

 

求凸包的最小外接矩形的面积。

思路：

旋转卡壳
给定点集S,求S的最小覆盖矩形
最小覆盖矩形的四条边上，其中一条边有至少两个点，其他边上至少有一个点。
然后沿着凸包的边旋转，维护矩形另外三条边上的点。

#include<stdio.h>
#include<cmath>
#include<algorithm>
#define eps 1e-8
#define N 50010
using namespace std;

struct Point
{
    double x,y;
    Point(){}
    Point(double x0,double y0):x(x0),y(y0){}
};

Point p[N];
int con[N];
int cn;
int n;

struct Line
{
    Point a,b;
    Line(){}
    Line(Point a0,Point b0):a(a0),b(b0){}
};

double Xmult(Point o,Point a,Point b)
{
    return (a.x-o.x)*(b.y-o.y)-(b.x-o.x)*(a.y-o.y);
}
double Dmult(Point o,Point a,Point b)
{
    return (a.x-o.x)*(b.x-o.x)+(a.y-o.y)*(b.y-o.y);
}

int Sig(double a)
{
    return a<-eps?-1:a>eps;
}

double Dis(Point a,Point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

int cmp(Point a,Point b)
{
    double d=Xmult(p[0],a,b);
    if(d>0)
        return 1;
    if(d==0 && Dis(p[0],a)<Dis(p[0],b))
        return 1;
    return 0;
}

double min(double a,double b)
{
    return a<b?a:b;
}

void Graham()
{
    int i,ind=0;
    for(i=1;i<n;i++)
    if(p[ind].y>p[i].y || (p[ind].y==p[i].y) && p[ind].x>p[i].x)
         ind=i;
    swap(p[ind],p[0]);
    sort(p+1,p+n,cmp);
    con[0]=0;
    con[1]=1;
    cn=1;
    for(i=2;i<n;i++)
    {
         while(cn>0 && Sig(Xmult(p[con[cn-1]],p[con[cn]],p[i]))<=0)
         cn--;
         con[++cn]=i;
    }
    int tmp=cn;
    for(i=n-2;i>=0;i--)
    {
         while(cn>tmp && Sig(Xmult(p[con[cn-1]],p[con[cn]],p[i]))<=0)
         cn--;
         con[++cn]=i;
    }
}

double Solve()
{
    int t,r,l;
    double ans=999999999;
    t=r=1;
    if(cn<3)
        return 0;
    for(int i=0;i<cn;i++)
    {
        while(Sig( Xmult(p[con[i]],p[con[i+1]],p[con[t+1]])-
             Xmult(p[con[i]],p[con[i+1]],p[con[t]])   )>0)
        t=(t+1)%cn;
        while(Sig( Dmult(p[con[i]],p[con[i+1]],p[con[r+1]])-
             Dmult(p[con[i]],p[con[i+1]],p[con[r]])   )>0)
        r=(r+1)%cn;
        if(!i) l=r;
        while(Sig( Dmult(p[con[i]],p[con[i+1]],p[con[l+1]])-
             Dmult(p[con[i]],p[con[i+1]],p[con[l]])   )<=0)
        l=(l+1)%cn;
        double d=Dis(p[con[i]],p[con[i+1]]);
        double tmp=Xmult(p[con[i]],p[con[i+1]],p[con[t]])*
      ( Dmult(p[con[i]],p[con[i+1]],p[con[r]])-
        Dmult(p[con[i]],p[con[i+1]],p[con[l]]) )/d/d;
        ans=min(ans,tmp);
    }
    return ans;
}

int main()
{
    int i;
    while(scanf("%d",&n) && n)
    {
         for(i=0;i<n;i++)
             scanf("%lf%lf",&p[i].x,&p[i].y);
         Graham();
         printf("%.4f\n",Solve());
    }
    return 0;
}