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;
}