uva10005(Packing polygons)

很簡單，裸的最小圓覆蓋。同 zoj 1450.

10005 - Packing polygons

Time limit: 3.000 seconds

Packing polygons

Given a polygon of n points (not necessarily convex), your goal is to say whether there is a circle of a given a radiusR that contains the polygon or not.

Input

The input consists of several input cases. The first line of each input case is the numbern (with n < 100) of vertices in the polygon. Then you are givenn lines each containing a couple of integers that define the coordinates of the vertices. The last line of the input case will be a real number indicating the radiusR of the circle.

The end of the input will be signaled by an input case with n = 0 vertices, that must not be processed.

You may assume that no vertex will appear twice in any given input case.

Output

If the polygon can be packed in a circle of the given radius you have to print:

The polygon can be packed in the circle.

If the polygon cannot be packed you have to print:

There is no way of packing that polygon.

Sample Input

3
0 0
1 0
0 1
1.0
3
0 0
1 0
0 1
0.1
0

Sample Output

The polygon can be packed in the circle.
There is no way of packing that polygon.

#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstring>
#define maxn 1005
using namespace std; 

struct tpoint
{
	double x,y;
	tpoint operator -(const tpoint &a) const
	{
		tpoint p1;
		p1.x=x-a.x;
		p1.y=y-a.y;
		return p1;
	} 
};
struct tcircle
{
	double r;
	tpoint centre;
};
struct ttriangle
{
	tpoint t[3];
};
tcircle c;
tpoint a[maxn];
double distances(tpoint p1,tpoint p2)
{
	tpoint p3;
	p3.x=p2.x-p1.x;
	p3.y=p2.y-p1.y;
	return sqrt(p3.x*p3.x+p3.y*p3.y);
}
double trianglearea(ttriangle t)
{
	tpoint p1,p2;
	p1=t.t[1]-t.t[0];
	p2=t.t[2]-t.t[0];
	return fabs(p1.x*p2.y-p1.y*p2.x)/2;
}
tcircle circumcircleoftriangle(ttriangle t)
{
	tcircle tmp;
	double a,b,c,c1,c2;
	double xa,xb,xc,ya,yb,yc;
	a = distances( t.t[0] , t.t[1] );
	b=distances(t.t[1],t.t[2]);
	c=distances(t.t[2],t.t[0]);
	tmp.r=a*b*c/trianglearea(t)/4;
	xa=t.t[0].x;ya=t.t[0].y;
	xb=t.t[1].x;yb=t.t[1].y;
	xc=t.t[2].x;yc=t.t[2].y;
	c1=(xa*xa+ya*ya-xb*xb-yb*yb)/2;
	c2=(xa*xa+ya*ya-xc*xc-yc*yc)/2;
	tmp.centre.x=(c1*(ya-yc)-c2*(ya-yb))/
	((xa-xb)*(ya-yc)-(xa-xc)*(ya-yb));
	tmp.centre.y=(c1*(xa-xc)-c2*(xa-xb))/
	((ya-yb)*(xa-xc)-(ya-yc)*(xa-xb));
	return tmp;
}
tcircle mincircle2(int tce,ttriangle ce)
{
	tcircle tmp;
	if(tce == 0) tmp.r=-2;
	else if(tce == 1)
	{
		tmp.centre=ce.t[0];
		tmp.r=0;
	}
	else if(tce == 2)
	{
		tmp.r=distances(ce.t[0],ce.t[1])/2;
		tmp.centre.x=(ce.t[0].x+ce.t[1].x)/2;
		tmp.centre.y=(ce.t[0].y+ce.t[1].y)/2;
	}
	else if(tce == 3)
	tmp=circumcircleoftriangle(ce);
	return tmp;
}
void mincircle(int t,int tce,ttriangle ce)
{
	int i,j;
	tpoint tmp;
	c=mincircle2(tce,ce);
	if(tce == 3) return ;
	for(i=1;i<=t;i++)
	{
		if(distances(a[i],c.centre)>c.r)
		{
			ce.t[tce]=a[i];
			mincircle(i-1,tce+1,ce);
			tmp=a[i];
			for(j = i;j>=2;j--)
			{
				a[j]=a[j-1];
			}
			a[1]=tmp;
		}
	}
}
void run(int n)
{
	ttriangle ce;
	int i;
	mincircle(n,0,ce);
//	printf("%.2lf %.2lf %.2lf\n",c.centre.x,c.centre.y,c.r);
}
int main()
{
	int n;
	double rr;
	while(scanf("%d",&n)!=EOF && n)
	{
		for(int i=1;i<=n;i++)
		scanf("%lf%lf",&a[i].x,&a[i].y);
		cin>>rr;
		run(n);
		if(c.r<=rr) cout<<"The polygon can be packed in the circle."<<endl;
		else cout<<"There is no way of packing that polygon."<<endl;
	}
	return 0;
}