Poj 3525 Most Distant Point from the Sea

地址：http://poj.org/problem?id=3525

题目：

Description

The main land of Japan called Honshu is an island surrounded by the sea. In such an island, it is natural to ask a question: “Where is the most distant point from the sea?” The answer to this question for Honshu was found in 1996. The most distant point is located in former Usuda Town, Nagano Prefecture, whose distance from the sea is 114.86 km.

In this problem, you are asked to write a program which, given a map of an island, finds the most distant point from the sea in the island, and reports its distance from the sea. In order to simplify the problem, we only consider maps representable by convex polygons.

Input

The input consists of multiple datasets. Each dataset represents a map of an island, which is a convex polygon. The format of a dataset is as follows.

Every input item in a dataset is a non-negative integer. Two input items in a line are separated by a space.

n in the first line is the number of vertices of the polygon, satisfying 3 ≤ n ≤ 100. Subsequent n lines are the x- and y-coordinates of the n vertices. Line segments (xi, yi)–(xi+1, yi+1) (1 ≤ i ≤ n − 1) and the line segment (xn, yn)–(x1, y1) form the border of the polygon in counterclockwise order. That is, these line segments see the inside of the polygon in the left of their directions. All coordinate values are between 0 and 10000, inclusive.

You can assume that the polygon is simple, that is, its border never crosses or touches itself. As stated above, the given polygon is always a convex one.

The last dataset is followed by a line containing a single zero.

Output

For each dataset in the input, one line containing the distance of the most distant point from the sea should be output. An output line should not contain extra characters such as spaces. The answer should not have an error greater than 0.00001 (10−5). You may output any number of digits after the decimal point, provided that the above accuracy condition is satisfied.

Sample Input

4
0 0
10000 0
10000 10000
0 10000
3
0 0
10000 0
7000 1000
6
0 40
100 20
250 40
250 70
100 90
0 70
3
0 0
10000 10000
5000 5001
0

Sample Output

5000.000000
494.233641
34.542948
0.353553
思路：二分距离d，然后将所有边内推距离d，之后用半平面交判断即可。
　　至于怎么内推，找出线段ab的方向向量（x,y）,那么向量c(-y,x)即是ab的垂线，再将c单位化后乘以d，线段(a+c)(b+c)即是内推后的线段位置。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>

using namespace std;
const double eps = 1e-10;
//点
class Point
{
public:
double x, y;

Point(){}
Point(double x, double y):x(x),y(y){}

bool operator < (const Point &_se) const
{
return x<_se.x || (x==_se.x && y<_se.y);
}
/*******判断ta与tb的大小关系*******/
static int sgn(double ta,double tb)
{
if(fabs(ta-tb)<eps)return 0;
if(ta<tb)   return -1;
return 1;
}
static double xmult(const Point &po, const Point &ps, const Point &pe)
{
return (ps.x - po.x) * (pe.y - po.y) - (pe.x - po.x) * (ps.y - po.y);
}
friend Point operator + (const Point &_st,const Point &_se)
{
return Point(_st.x + _se.x, _st.y + _se.y);
}
friend Point operator - (const Point &_st,const Point &_se)
{
return Point(_st.x - _se.x, _st.y - _se.y);
}
//点位置相同(double类型)
bool operator == (const Point &_off) const
{
return  Point::sgn(x, _off.x) == 0 && Point::sgn(y, _off.y) == 0;
}
//点位置不同(double类型)
bool operator != (const Point &_Off) const
{
return ((*this) == _Off) == false;
}
//两点间距离的平方
static double dis2(const Point &_st,const Point &_se)
{
return (_st.x - _se.x) * (_st.x - _se.x) + (_st.y - _se.y) * (_st.y - _se.y);
}
//两点间距离
static double dis(const Point &_st, const Point &_se)
{
return sqrt((_st.x - _se.x) * (_st.x - _se.x) + (_st.y - _se.y) * (_st.y - _se.y));
}
};
//两点表示的向量
class Line
{
public:
Point s, e;//两点表示，起点[s]，终点[e]
double a, b, c;//一般式,ax+by+c=0

Line(){}
Line(const Point &s, const Point &e):s(s),e(e){}
Line(double _a,double _b,double _c):a(_a),b(_b),c(_c){}

//向量与点的叉乘,参数：点[_Off]
//[点相对向量位置判断]
double operator /(const Point &_Off) const
{
return (_Off.y - s.y) * (e.x - s.x) - (_Off.x - s.x) * (e.y - s.y);
}
//向量与向量的叉乘,参数：向量[_Off]
friend double operator /(const Line &_st,const Line &_se)
{
return (_st.e.x - _st.s.x) * (_se.e.y - _se.s.y) - (_st.e.y - _st.s.y) * (_se.e.x - _se.s.x);
}
friend double operator *(const Line &_st,const Line &_se)
{
return (_st.e.x - _st.s.x) * (_se.e.x - _se.s.x) - (_st.e.y - _st.s.y) * (_se.e.y - _se.s.y);
}
//从两点表示转换为一般表示
//a=y2-y1,b=x1-x2,c=x2*y1-x1*y2
bool pton()
{
a = e.y - s.y;
b = s.x - e.x;
c = e.x * s.y - e.y * s.x;
return true;
}

//-----------点和直线（向量）-----------
//点在向量左边（右边的小于号改成大于号即可,在对应直线上则加上=号）
//参数：点[_Off],向量[_Ori]
friend bool operator<(const Point &_Off, const Line &_Ori)
{
return (_Ori.e.y - _Ori.s.y) * (_Off.x - _Ori.s.x)
< (_Off.y - _Ori.s.y) * (_Ori.e.x - _Ori.s.x);
}

//点在直线上,参数：点[_Off]
bool lhas(const Point &_Off) const
{
return Point::sgn((*this) / _Off, 0) == 0;
}
//点在线段上,参数：点[_Off]
bool shas(const Point &_Off) const
{
return lhas(_Off)
&& Point::sgn(_Off.x - min(s.x, e.x), 0) > 0 && Point::sgn(_Off.x - max(s.x, e.x), 0) < 0
&& Point::sgn(_Off.y - min(s.y, e.y), 0) > 0 && Point::sgn(_Off.y - max(s.y, e.y), 0) < 0;
}

//点到直线/线段的距离
//参数： 点[_Off], 是否是线段[isSegment](默认为直线)
double dis(const Point &_Off, bool isSegment = false)
{
///化为一般式
pton();

//到直线垂足的距离
double td = (a * _Off.x + b * _Off.y + c) / sqrt(a * a + b * b);

//如果是线段判断垂足
if(isSegment)
{
double xp = (b * b * _Off.x - a * b * _Off.y - a * c) / ( a * a + b * b);
double yp = (-a * b * _Off.x + a * a * _Off.y - b * c) / (a * a + b * b);
double xb = max(s.x, e.x);
double yb = max(s.y, e.y);
double xs = s.x + e.x - xb;
double ys = s.y + e.y - yb;
if(xp > xb + eps || xp < xs - eps || yp > yb + eps || yp < ys - eps)
td = min(Point::dis(_Off,s), Point::dis(_Off,e));
}

return fabs(td);
}

//关于直线对称的点
Point mirror(const Point &_Off) const
{
///注意先转为一般式
Point ret;
double d = a * a + b * b;
ret.x = (b * b * _Off.x - a * a * _Off.x - 2 * a * b * _Off.y - 2 * a * c) / d;
ret.y = (a * a * _Off.y - b * b * _Off.y - 2 * a * b * _Off.x - 2 * b * c) / d;
return ret;
}
//计算两点的中垂线
static Line ppline(const Point &_a, const Point &_b)
{
Line ret;
ret.s.x = (_a.x + _b.x) / 2;
ret.s.y = (_a.y + _b.y) / 2;
//一般式
ret.a = _b.x - _a.x;
ret.b = _b.y - _a.y;
ret.c = (_a.y - _b.y) * ret.s.y + (_a.x - _b.x) * ret.s.x;
//两点式
if(std::fabs(ret.a) > eps)
{
ret.e.y = 0.0;
ret.e.x = - ret.c / ret.a;
if(ret.e == ret. s)
{
ret.e.y = 1e10;
ret.e.x = - (ret.c - ret.b * ret.e.y) / ret.a;
}
}
else
{
ret.e.x = 0.0;
ret.e.y = - ret.c / ret.b;
if(ret.e == ret. s)
{
ret.e.x = 1e10;
ret.e.y = - (ret.c - ret.a * ret.e.x) / ret.b;
}
}
return ret;
}

//------------直线和直线（向量）-------------
//直线重合,参数：直线向量[_st],[_se]
static bool equal(const Line &_st, const Line &_se)
{
return _st.lhas(_se.e) && _se.lhas(_se.s);
}
//直线平行，参数：直线向量[_st],[_se]
static bool parallel(const Line &_st,const Line &_se)
{
return Point::sgn(_st / _se, 0) == 0;
}
//两直线（线段）交点，参数：直线向量[_st],[_se],交点
//返回-1代表平行，0代表重合，1代表相交
static bool crossLPt(const Line &_st,const Line &_se,Point &ret)
{
if(Line::parallel(_st,_se))
{
if(Line::equal(_st,_se)) return 0;
return -1;
}
ret = _st.s;
double t = (Line(_st.s,_se.s)/_se)/(_st/_se);
ret.x += (_st.e.x - _st.s.x) * t;
ret.y += (_st.e.y - _st.s.y) * t;
return 1;
}
//------------线段和直线（向量）----------
//线段和直线交
//参数：直线[_st],线段[_se]
friend bool crossSL(const Line &_st,const Line &_se)
{
return Point::sgn((_st / _se.s) * (_st / _se.e) ,0) <= 0;
}

//------------线段和线段（向量）----------
//判断线段是否相交(注意添加eps)，参数：线段[_st],线段[_se]
static bool isCrossSS(const Line &_st,const Line &_se)
{
//1.快速排斥试验判断以两条线段为对角线的两个矩形是否相交
//2.跨立试验（等于0时端点重合）
return
max(_st.s.x, _st.e.x) >= min(_se.s.x, _se.e.x) &&
max(_se.s.x, _se.e.x) >= min(_st.s.x, _st.e.x) &&
max(_st.s.y, _st.e.y) >= min(_se.s.y, _se.e.y) &&
max(_se.s.y, _se.e.y) >= min(_st.s.y, _st.e.y) &&
Point::sgn((_st / Line(_st.s, _se.s)) * (_st / Line(_st.s, _se.e)), 0) <= 0 &&
Point::sgn((_se / Line(_se.s, _st.s)) * (_se / Line(_se.s, _st.e)), 0) <= 0;
}
};
class Polygon
{
public:
const static int maxpn = 100;
Point pt[maxpn];//点（顺时针或逆时针）
int n;//点的个数

Point& operator[](int _p)
{
return pt[_p];
}

//求多边形面积，多边形内点必须顺时针或逆时针
double area() const
{
double ans = 0.0;
for(int i = 0; i < n; i ++)
{
int nt = (i + 1) % n;
ans += pt[i].x * pt[nt].y - pt[nt].x * pt[i].y;
}
return fabs(ans / 2.0);
}
//求多边形重心，多边形内点必须顺时针或逆时针
Point gravity() const
{
Point ans;
ans.x = ans.y = 0.0;
double area = 0.0;
for(int i = 0; i < n; i ++)
{
int nt = (i + 1) % n;
double tp = pt[i].x * pt[nt].y - pt[nt].x * pt[i].y;
area += tp;
ans.x += tp * (pt[i].x + pt[nt].x);
ans.y += tp * (pt[i].y + pt[nt].y);
}
ans.x /= 3 * area;
ans.y /= 3 * area;
return ans;
}
//判断点在凸多边形内，参数：点[_Off]
bool chas(const Point &_Off) const
{
double tp = 0, np;
for(int i = 0; i < n; i ++)
{
np = Line(pt[i], pt[(i + 1) % n]) / _Off;
if(tp * np < -eps)
return false;
tp = (fabs(np) > eps)?np: tp;
}
return true;
}
//判断点是否在任意多边形内[射线法]，O(n)
bool ahas(const Point &_Off) const
{
int ret = 0;
double infv = 1e-10;//坐标系最大范围
Line l = Line(_Off, Point( -infv ,_Off.y));
for(int i = 0; i < n; i ++)
{
Line ln = Line(pt[i], pt[(i + 1) % n]);
if(fabs(ln.s.y - ln.e.y) > eps)
{
Point tp = (ln.s.y > ln.e.y)? ln.s: ln.e;
if(fabs(tp.y - _Off.y) < eps && tp.x < _Off.x + eps)
ret ++;
}
else if(Line::isCrossSS(ln,l))
ret ++;
}
return (ret % 2 == 1);
}
//凸多边形被直线分割,参数：直线[_Off]
Polygon split(Line _Off)
{
//注意确保多边形能被分割
Polygon ret;
Point spt[2];
double tp = 0.0, np;
bool flag = true;
int i, pn = 0, spn = 0;
for(i = 0; i < n; i ++)
{
if(flag)
pt[pn ++] = pt[i];
else
ret.pt[ret.n ++] = pt[i];
np = _Off / pt[(i + 1) % n];
if(tp * np < -eps)
{
flag = !flag;
Line::crossLPt(_Off,Line(pt[i], pt[(i + 1) % n]),spt[spn++]);
}
tp = (fabs(np) > eps)?np: tp;
}
ret.pt[ret.n ++] = spt[0];
ret.pt[ret.n ++] = spt[1];
n = pn;
return ret;
}

/** 卷包裹法求点集凸包，_p为输入点集，_n为点的数量 **/
void ConvexClosure(Point _p[],int _n)
{
sort(_p,_p+_n);
n=0;
for(int i=0;i<_n;i++)
{
while(n>1&&Point::sgn(Line(pt[n-2],pt[n-1])/Line(pt[n-2],_p[i]),0)<=0)
n--;
pt[n++]=_p[i];
}
int _key=n;
for(int i=_n-2;i>=0;i--)
{
while(n>_key&&Point::sgn(Line(pt[n-2],pt[n-1])/Line(pt[n-2],_p[i]),0)<=0)
n--;
pt[n++]=_p[i];
}
if(n>1)   n--;//除去重复的点，该点已是凸包凸包起点
}
//    /****** 寻找凸包的graham 扫描法********************/
//    /****** _p为输入的点集,_n为点的数量****************/
//    /**使用时需把gmp函数放在类外，并且看情况修改pt[0]**/
//    bool gcmp(const Point &ta,const Point &tb)/// 选取与最后一条确定边夹角最小的点，即余弦值最大者
//    {
//        double tmp=Line(pt[0],ta)/Line(pt[0],tb);
//        if(Point::sgn(tmp,0)==0)
//            return Point::dis(pt[0],ta)<Point::dis(pt[0],tb);
//        else if(tmp>0)
//            return 1;
//        return 0;
//    }
//    void graham(Point _p[],int _n)
//    {
//        int cur=0;
//        for(int i=1;i<_n;i++)
//            if(Point::sgn(_p[cur].y,_p[i].y)>0 || (Point::sgn(_p[cur].y,_p[i].y)==0 && Point::sgn(_p[cur].x,_p[i].x)>0))
//                cur=i;
//        swap(_p[cur],_p[0]);
//        n=0,pt[n++]=_p[0];
//        if(_n==1)   return;
//        sort(_p+1,_p+_n,Polygon::gcmp);
//        pt[n++]=_p[1],pt[n++]=_p[2];
//        for(int i=3;i<_n;i++)
//        {
//            while(Point::sgn(Line(pt[n-2],pt[n-1])/Line(pt[n-2],_p[i]),0)<0)
//                n--;
//            pt[n++]=_p[i];
//        }
//    }
//凸包旋转卡壳(注意点必须顺时针或逆时针排列)
//返回值凸包直径的平方（最远两点距离的平方）
double rotating_calipers()
{
int i = 1;
double ret = 0.0;
pt
 = pt[0];
for(int j = 0; j < n; j ++)
{
while(fabs(Point::xmult(pt[i+1],pt[j], pt[j + 1])) > fabs(Point::xmult(pt[i],pt[j], pt[j + 1])) + eps)
i = (i + 1) % n;
//pt[i]和pt[j],pt[i + 1]和pt[j + 1]可能是对踵点
ret = (ret, max(Point::dis(pt[i],pt[j]), Point::dis(pt[i + 1],pt[j + 1])));
}
return ret;
}

//凸包旋转卡壳(注意点必须逆时针排列)
//返回值两凸包的最短距离
double rotating_calipers(Polygon &_Off)
{
int i = 0;
double ret = 1e10;//inf
pt
 = pt[0];
_Off.pt[_Off.n] = _Off.pt[0];
//注意凸包必须逆时针排列且pt[0]是左下角点的位置
while(_Off.pt[i + 1].y > _Off.pt[i].y)
i = (i + 1) % _Off.n;
for(int j = 0; j < n; j ++)
{
double tp;
//逆时针时为 >,顺时针则相反
while((tp = Point::xmult(_Off.pt[i + 1],pt[j], pt[j + 1]) - Point::xmult(_Off.pt[i], pt[j], pt[j + 1])) > eps)
i = (i + 1) % _Off.n;
//(pt[i],pt[i+1])和(_Off.pt[j],_Off.pt[j + 1])可能是最近线段
ret = min(ret, Line(pt[j], pt[j + 1]).dis(_Off.pt[i], true));
ret = min(ret, Line(_Off.pt[i], _Off.pt[i + 1]).dis(pt[j + 1], true));
if(tp > -eps)//如果不考虑TLE问题最好不要加这个判断
{
ret = min(ret, Line(pt[j], pt[j + 1]).dis(_Off.pt[i + 1], true));
ret = min(ret, Line(_Off.pt[i], _Off.pt[i + 1]).dis(pt[j], true));
}
}
return ret;
}

//-----------半平面交-------------
//复杂度:O(nlog2(n))
//#include <algorithm>
//半平面计算极角函数[如果考虑效率可以用成员变量记录]
static double hpc_pa(const Line &_Off)
{
return atan2(_Off.e.y - _Off.s.y, _Off.e.x - _Off.s.x);
}
//半平面交排序函数[优先顺序: 1.极角 2.前面的直线在后面的左边]
static bool hpc_cmp(const Line &l, const Line &r)
{
double lp = hpc_pa(l), rp = hpc_pa(r);
if(fabs(lp - rp) > eps)
return lp < rp;
return Point::xmult(r.s,l.s, r.e) < -eps;
}
static int judege(const Line &_lx,const Line &_ly,const Line &_lz)
{
Point tmp;
Line::crossLPt(_lx,_ly,tmp);
return Point::sgn(Point::xmult(_lz.s,tmp,_lz.e),0);
}
//获取半平面交的多边形（多边形的核）
//参数：向量集合[l]，向量数量[ln];(半平面方向在向量左边）
//函数运行后如果n[即返回多边形的点数量]为0则不存在半平面交的多边形（不存在区域或区域面积无穷大）
Polygon& halfPanelCross(Line _Off[], int ln)
{
Line dequeue[maxpn];//用于计算的双端队列
int i, tn, bot, top;
sort(_Off, _Off + ln, hpc_cmp);
//平面在向量左边的筛选
for(i = tn = 1; i < ln; i ++)
if(fabs(hpc_pa(_Off[i]) - hpc_pa(_Off[i - 1])) > eps)
_Off[tn ++] = _Off[i];
ln = tn, n = 0, bot = 0, top = 1;
dequeue[0] = _Off[0];
dequeue[1] = _Off[1];
for(i = 2; i < ln; i ++)
{
while(bot < top &&  Polygon::judege(dequeue[top],dequeue[top-1],_Off[i]) > 0)
top --;
while(bot < top &&  Polygon::judege(dequeue[bot],dequeue[bot+1],_Off[i]) > 0)
bot ++;
dequeue[++ top] = _Off[i];
}
while(bot < top && Polygon::judege(dequeue[top],dequeue[top-1],dequeue[bot]) > 0)
top --;
while(bot < top && Polygon::judege(dequeue[bot],dequeue[bot+1],dequeue[top]) > 0)
bot ++;
//计算交点(注意不同直线形成的交点可能重合)
if(top <= bot + 1)
return (*this);
for(i = bot; i < top; i ++)
Line::crossLPt(dequeue[i],dequeue[i + 1],pt[n++]);
if(bot < top + 1)
Line::crossLPt(dequeue[bot],dequeue[top],pt[n++]);
return (*this);
}
};

Point pt[100];
Line ln[100],ls[100];
Polygon py;
int n;

void ml(Line &lx,Line &ly,double t)
{
Point tmp=lx.e-lx.s,of;
of=Point(-tmp.y,tmp.x);
double dis=sqrt(of.x*of.x+of.y*of.y);
of.x=of.x*t/dis,of.y=of.y*t/dis;
ly.s=lx.s+of,ly.e=lx.e+of;
}
int sc(double x)
{
for(int i=0;i<n;i++)
ml(ln[i],ls[i],x);
//for(int i=0;i<n;i++)
//    printf("%.2f %.2f %.2f %.2f\n",ls[i].s.x,ls[i].s.y,ls[i].e.x,ls[i].e.y);
py.halfPanelCross(ls,n);
//printf("==%d\n",py.n);
//for(int i=0;i<py.n;i++)
//    printf("%.2f %.2f\n",py.pt[i].x,py.pt[i].y);
return py.n;
}
int main(void)
{
while(scanf("%d",&n)&&n)
{
for(int i=0;i<n;i++)
scanf("%lf%lf",&pt[i].x,&pt[i].y);
pt
=pt[0];
for(int i=0;i<n;i++)
ln[i]=Line(pt[i],pt[i+1]);
//for(int i=0;i<n;i++)
//    printf("%.2f %.2f %.2f %.2f\n",ln[i].s.x,ln[i].s.y,ln[i].e.x,ln[i].e.y);
double l=0,r=1e8;
while(fabs(r-l)>=1e-5)
{
double mid=(r+l)/2.0;
if(sc(mid)) l=mid;
else    r=mid;
}
printf("%.6f\n",r);
}
return 0;
}