poj 3525Most Distant Point from the Sea【二分+半平面交】

相当于多边形内最大圆，二分半径r，然后把每条边内收r，求是否有半平面交（即是否合法）

#include<iostream>

#include<cstdio>

#include<algorithm>

#include<cmath>

using namespace std;

const int N=205;

const double eps=1e-6;

int n;

struct dian

{

	double x,y;

	dian(double X=0,double Y=0)

	{

		x=X,y=Y;

	}

	dian operator + (const dian &a)

	{

		return dian(x+a.x,y+a.y);

	}

	dian operator - (const dian &a)

	{

		return dian(x-a.x,y-a.y);

	}

	dian operator * (const double &a) const

	{

		return dian(x*a,y*a);

	}

	dian operator / (const double &a) const

	{

		return dian(x/a,y/a);

	}

}p[N];

struct bian

{

	dian s,v;

	bian(dian S=dian(),dian V=dian())

	{

		s=S,v=V;

	}

}b[N],l[N],s[N];

int read()

{

	int r=0,f=1;

	char p=getchar();

	while(p>'9'||p<'0')

	{

		if(p=='-')

			f=-1;

		p=getchar();

	}

	while(p>='0'&&p<='9')

	{

		r=r*10+p-48;

		p=getchar();

	}

	return r*f;

}

double cj(dian a,dian b)

{

	return a.x*b.y-a.y*b.x;

}

double mj(dian a,dian b,dian c)

{

	return cj(b-a,c-a)/2;

}

dian jd(bian x,bian y)

{

	return x.s+x.v*(cj(x.s-y.s,y.v)/cj(y.v,x.v));

}

bool px(bian a,bian b)

{

	return cj(a.v,b.v)==0;

}

bool bn(bian a,bian b)

{

	int ar=cj(a.v,b.v);

	return ar>0||(ar==0&&cj(a.v,b.s-a.s)>0);

}

bool dn(dian x,bian y)

{

	return cj(y.v,x-y.s)<=0;

}

bool cmp(const bian &x,const bian &y)

{

	if(x.v.y==0&&y.v.y==0)

		return x.v.x<y.v.x;

	if((x.v.y<=0)==(y.v.y<=0))

		return bn(x,y);

	return x.v.y<y.v.y;

}

double dis2(bian a)

{

	return sqrt(a.v.x*a.v.x+a.v.y*a.v.y);

}

bian yi(bian a,double r)

{

	return bian(dian(a.s.x-a.v.y*r/dis2(a),a.s.y+a.v.x*r/dis2(a)),a.v);

}

bool ok(double r)

{

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

		l[i]=yi(b[i],r);

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

	int top=0;

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

		if(i==1||!px(l[i],l[i-1]))

			l[++top]=l[i];

	n=top;

	int ll=1,rr=2;

	s[1]=l[1],s[2]=l[2];

	for(int i=3;i<=n;i++)

	{

		while(ll<rr&&dn(jd(s[rr],s[rr-1]),l[i]))

			rr--;

		while(ll<rr&&dn(jd(s[ll],s[ll+1]),l[i]))

			ll++;

		s[++rr]=l[i];

	}

	while(ll<rr&&dn(jd(s[rr],s[rr-1]),s[ll]))

		rr--;

	return rr-ll>1;

}

int main()

{

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

	{

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

			p[i].x=read(),p[i].y=read();

		p[n+1]=p[1];

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

			b[i]=bian(p[i],p[i+1]-p[i]);

		double l=0,r=10005,ans=0;

		while(r-l>=eps)

		{

			double mid=(l+r)/2;//printf("%.6f\n",mid);

			if(ok(mid))

				l=mid,ans=mid;

			else

				r=mid;

		}

		printf("%.6f\n",ans);

	}

	return 0;

}

巴特西

poj 3525Most Distant Point from the Sea【二分+半平面交】

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