题目大意：给出一个岛的海岸线的轮廓，求这个岛上的所有点到海岸的最长距离是多少。

思路：求多边形内切圆的问题要用二分+半平面交解决。二分半径的长度，然后将所有的边向左侧移动这个二分的长度，然后利用半平面交来判断是否能够满足条件。如果满足条件就提高下界，否则减小上界。

我的移动的方法是这样的，首先每条边都要用点向量式来表示，就是边上任意一点和这条边的方向向量。这样做以后的操作会方便很多。然后将每个直线的向左的法向量求出来（比如l的向量是v(x,y)，那么它向左侧的法向量是(-y,x)），然后将法向量化成单位向量，然后这个直线方向向量不变，直线上的点向法向量方向移动二分的长度。详见代码。

CODE：

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 110
#define EPS 1e-10
#define DCMP(a) (fabs(a) < EPS)
using namespace std;

struct Point{
	double x,y;

	Point(double _ = .0,double __ = .0):x(_),y(__) {}
	Point operator +(const Point &a)const {
		return Point(x + a.x,y + a.y);
	}
	Point operator -(const Point &a)const {
		return Point(x - a.x,y - a.y);
	}
	Point operator *(double a)const {
		return Point(x * a,y * a);
	}
	void Read() {
		scanf("%lf%lf",&x,&y);
	}
}point[MAX],p[MAX];
struct Line{
	Point p,v;
	double alpha;

	Line(Point _,Point __):p(_),v(__) {
		alpha = atan2(v.y,v.x);
	}
	Line() {}
	bool operator <(const Line &a)const {
		return alpha < a.alpha;
	}
}line[MAX],q[MAX];

int points,lines;

inline double Cross(const Point &a,const Point &b)
{
	return a.x * b.y - a.y * b.x;
}

inline bool OnLeft(const Line &l,const Point &p)
{
	return Cross(l.v,p - l.p) >= 0;
}

inline double Calc(const Point &a,const Point &b)
{
	return sqrt((a.x - b.x) * (a.x - b.x) + 
				(a.y - b.y) * (a.y - b.y));
}

inline Point GetIntersection(const Line &a,const Line &b)
{
	Point u = a.p - b.p;
	double temp = Cross(b.v,u) / Cross(a.v,b.v);
	return a.p + a.v * temp;
}

inline bool HalfPlaneIntersection()
{
	int front = 1,tail = 1;
	q[tail] = line[1];
	for(int i = 2;i <= lines; ++i) {
		while(front < tail && !OnLeft(line[i],p[tail - 1]))	--tail;
		while(front < tail && !OnLeft(line[i],p[front]))	++front;
		if(DCMP(Cross(line[i].v,q[tail].v)))
			q[tail] = OnLeft(line[i],q[tail].p) ? q[tail]:line[i];
		else	q[++tail] = line[i];
		if(front < tail)	p[tail - 1] = GetIntersection(q[tail],q[tail - 1]);
	}
	while(front < tail && !OnLeft(line[front],p[tail - 1]))	--tail;
	if(tail - front <= 1)	return false;
	return tail > front;
}

inline void MakeLine(const Point &a,const Point &b,double adjustment)
{
	Point p = a,v = b - a;
	double length = Calc(a,b);
	Point _v(-v.y / length,v.x / length);
	p = p + _v * adjustment;
	line[++lines] = Line(p,v);
}

inline bool Judge(double ans)
{
	lines = 0;
	for(int i = 1;i < points; ++i)
		MakeLine(point[i],point[i + 1],ans);
	MakeLine(point[points],point[1],ans);
	sort(line + 1,line + lines + 1);
	return HalfPlaneIntersection();
}

int main()
{
	while(scanf("%d",&points),points) {
		for(int i = 1;i <= points; ++i)
			point[i].Read();
		double l = .0,r = 10000.0,ans = .0;
		while(r - l > EPS) {
			double mid = (l + r) * 0.5;
			if(Judge(mid))	l = mid,ans = mid;
			else	r = mid;
		}
		printf("%.6lf\n",ans);
	}
	return 0;
}

POJ 3525 Most Distant Point from the Sea 二分+半平面交