题意：凸多边形的小岛在海里，问岛上的点到海最远的距离。

分析：训练指南P279，二分答案，然后整个多边形往内部收缩，如果半平面交非空，那么这些点构成半平面，存在满足的点。

 

/************************************************* Author        :Running_Time* Created Time  :2015/11/10 星期二 14:16:17* File Name     :LA_3890.cpp ************************************************/#include 
     
      using namespace std;#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1 | 1typedef long long ll;const int N = 1e5 + 10;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;const double EPS = 1e-10;const double PI = acos (-1.0);int dcmp(double x)  {       //三态函数，减少精度问题    if (fabs (x) < EPS) return 0;    else    return x < 0 ? -1 : 1;}struct Point    {       //点的定义    double x, y;    Point () {}    Point (double x, double y) : x (x), y (y) {}    Point operator + (const Point &r) const {       //向量加法        return Point (x + r.x, y + r.y);    }    Point operator - (const Point &r) const {       //向量减法        return Point (x - r.x, y - r.y);    }    Point operator * (double p) const {       //向量乘以标量        return Point (x * p, y * p);    }    Point operator / (double p) const {       //向量除以标量        return Point (x / p, y / p);    }    bool operator < (const Point &r) const {       //点的坐标排序        return x < r.x || (x == r.x && y < r.y);    }    bool operator == (const Point &r) const {       //判断同一个点        return dcmp (x - r.x) == 0 && dcmp (y - r.y) == 0;    }};typedef Point Vector;       //向量的定义Point read_point(void)   {      //点的读入    double x, y;    scanf ("%lf%lf", &x, &y);    return Point (x, y);}double dot(Vector A, Vector B)  {       //向量点积    return A.x * B.x + A.y * B.y;}double cross(Vector A, Vector B)    {       //向量叉积    return A.x * B.y - A.y * B.x;}double length(Vector A) {       //向量长度，点积    return sqrt (dot (A, A));}double polar_angle(Vector A)  {     //向量极角    return atan2 (A.y, A.x);}Vector nomal(Vector A)  {       //向量的单位法向量    double len = length (A);    return Vector (-A.y / len, A.x / len);}struct Line {    Point p;    Vector v;    double ang;    Line () {}    Line (const Point &p, const Vector &v) : p (p), v (v)   {        ang = polar_angle (v);    }    bool operator < (const Line &r) const {        return ang < r.ang;    }    Point point(double a)   {        return p + v * a;    }};bool point_on_left(Point p, Line L) {    return cross (L.v, p - L.p) > 0;}Point line_line_inter(Point p, Vector V, Point q, Vector W)    {        //两直线交点，参数方程    Vector U = p - q;    double t = cross (W, U) / cross (V, W);    return p + V * t;}vector
      
        half_plane_inter(vector
       
         L)    {    sort (L.begin (), L.end ());    int first, last, n = L.size ();    Point *p = new Point[n];    Line *q = new Line[n];    q[first=last=0] = L[0];    for (int i=1; i
        
          ps;    if (last - first <= 1)  return ps;    p[last] = line_line_inter (q[last].p, q[last].v, q[first].p, q[first].v);    for (int i=first; i<=last; ++i) ps.push_back (p[i]);    return ps;}Point ps[110];Vector V[110], V2[110];int main(void)    {    int n;    while (scanf ("%d", &n) == 1)   {        if (!n) break;        for (int i=0; i
         
           EPS)    {            double mid = l + (r - l) / 2;            vector
          
            L; for (int i=0; i
           
             qs = half_plane_inter (L); int sz = qs.size (); if (sz == 0) r = mid; else l = mid; } printf ("%.6f\n", l); } //cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n"; return 0;}