旋转卡壳法求点集的最小覆盖矩形面积以及周长
uva12307
Problem H
Smallest Enclosing Rectangle
There are n points in 2D space. You're to find a smallest enclosing rectangle of these points. By "smallest" we mean either area or perimeter (yes, you have to solve both problems. The optimal rectangle for these two problems might be different). Note that the sides of the rectangle might not be parallel to the coordinate axes.
Input
There will be at most 10 test cases in the input. Each test case begins with a single integer n (3<=n<=100,000), the number of points. Each of the following n lines contains two real numbers x, y (-100,000<=x,y<=100,000), the coordinates of the points. The points will not be collinear. The last test case is followed by a line with n=0, which should not be processed.
Output
For each line, print the area of the minimum-area enclosing rectangle, and the perimeter of the minimum-perimeter enclosing rectangle, both rounded to two decimal places.
Sample Input
5 0 0 2 0 2 2 0 2 1 1 5 1 1 9 0 7 10 0 5 2 11 3 5 3 7 2 6 6 4 6 3 9 1 9 6 8 10 0
Output for the Sample Input
4.00 8.00 95.38 39.19 7.00 11.38 27.00 23.63程序：
#include"string.h"
#include"stdio.h"
#include"math.h"
#include"stdlib.h"
#define M 100009
#define inf 999999999
#define eps 1e-10
typedef struct node
{
double x,y,cos,dis;
}E;
E p[M],q[M],pp[M];
double max(double a,double b)
{
return a>b?a:b;
}
double min(double a,double b)
{
return a<b?a:b;
}
int cmp(const void *a,const void *b)
{
if(fabs((*(struct node*)a).cos-(*(struct node*)b).cos)<eps)
return (*(struct node*)a).dis>(*(struct node*)b).dis?1:-1;
else
return (*(struct node*)b).cos>(*(struct node*)a).cos?1:-1;
}
double angle(node p1,node p2)
{
double x1=p2.x-p1.x;
double y1=p2.y-p1.y;
double x2=1;
double y2=0;
return (x1*x2+y1*y2)/sqrt((x1*x1+y1*y1)*(x2*x2+y2*y2));
}
double pow(double x)
{
return x*x;
}
double Len(node p1,node p2)
{
return pow(p2.x-p1.x)+pow(p2.y-p1.y);
}
double cross(node p0,node p1,node p2)//叉积
{
double x1=p1.x-p0.x;
double y1=p1.y-p0.y;
double x2=p2.x-p0.x;
double y2=p2.y-p0.y;
return x1*y2-x2*y1;
}
double dot(node p0,node p1,node p2)//点积
{
double x1=p1.x-p0.x;
double y1=p1.y-p0.y;
double x2=p2.x-p1.x;
double y2=p2.y-p1.y;
return x1*x2+y1*y2;
}
int main()
{
int n,i,j;
while(scanf("%d",&n),n)
{
int tep;
E start;
start.x=start.y=inf;
for(i=0;i<n;i++)
{
scanf("%lf%lf",&p[i].x,&p[i].y);
if(p[i].y<start.y)
{
start=p[i];
tep=i;
}
else if(fabs(p[i].y-start.y)<eps)
{
if(p[i].x<start.x)
{
start=p[i];
tep=i;
}
}
}
p[tep].dis=0;
p[tep].cos=1.0;
for(i=0;i<n;i++)
{
if(i!=tep)
{
if(fabs(p[i].x-start.x)<eps&&fabs(p[i].y-start.y)<eps)
{
p[i].dis=0;
p[i].cos=1.0;
}
else
{
p[i].cos=angle(start,p[i]);
p[i].dis=Len(start,p[i]);
}
}
}
qsort(p,n,sizeof(p[0]),cmp);//按极角进行快排
int tt=0;
for(i=0;i<n;i++)//除重点
{
if(fabs(p[i].dis-p[(i+1)%n].dis)>eps||fabs(p[i].cos-p[(i+1)%n].cos)>eps)
pp[tt++]=p[i];
}
if(tt==0)
{
printf("0.00 0.00
");
continue;
}
if(tt==2)
{
printf("%.2lf %.2lf
",0.0,2*sqrt(Len(pp[0],pp[1])));
continue;
}
q[0]=pp[tt-1];
q[1]=pp[0];
q[2]=pp[1];
int cnt=2;
for(i=2;i<tt;i++)
{
while(cross(q[cnt-1],q[cnt],pp[i])<0)
{
cnt--;
}
q[++cnt]=pp[i];
}//求凸包
j=1;
int k1,k2;
k1=k2=1;
double S=inf,C=inf;
for(i=0;i<cnt;i++)
{
double w=sqrt(Len(q[i],q[(i+1)%cnt]));
while(cross(q[i],q[(i+1)%cnt],q[(j+1)%cnt])>cross(q[i],q[(i+1)%cnt],q[j%cnt]))
{
j++;
}
double high=cross(q[i],q[(i+1)%cnt],q[j%cnt])/w;
while(dot(q[i],q[(i+1)%cnt],q[(k1+1)%cnt])>dot(q[i],q[(i+1)%cnt],q[(k1%cnt)]))
{
k1++;
}
if(i==0)
k2=k1;
while(dot(q[i],q[(i+1)%cnt],q[(k2+1)%cnt])<=dot(q[i],q[(i+1)%cnt],q[(k2%cnt)]))
{
k2++;
}
double wide=(dot(q[i],q[(i+1)%cnt],q[(k1%cnt)])-dot(q[i],q[(i+1)%cnt],q[(k2%cnt)]))/w;
//printf("%.3lf %.3lf %.3lf
",high,wide,(high+wide)*2);
S=min(S,high*wide);
C=min(C,(high+wide)*2);
}
printf("%.2lf %.2lf
",S,C);
}
return 0;
}