LA 3890 (半平面交) Most Distant Point from the Sea

题意：

给出一个凸n边形，求多边形内部一点使得该点到边的最小距离最大。

分析：

最小值最大可以用二分。

多边形每条边的左边是一个半平面，将这n个半平面向左移动距离x，则将这个凸多边形缩小了。如果这n个半平面交非空，则存在这样距离为x的点，反之则不存在。

半平面交的代码还没有完全理解。

和凸包类似，先对这些半平面进行极角排序。每次新加入的平面可能让队尾的平面变得“无用”，从而需要删除。由于极角序是环形的，所以也可能让队首元素变得“无用”。所以用双端队列来维护。

 //#define LOCAL
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <cmath>
 #include <vector>
 using namespace std;

 const double eps = 1e-;

 struct Point
 {
     double x, y;
     Point(double x=, double y=):x(x), y(y){}
 };
 typedef Point Vector;
 Point operator + (Point A, Point B) { return Point(A.x+B.x, A.y+B.y); }
 Point operator - (Point A, Point B) { return Point(A.x-B.x, A.y-B.y); }
 Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); }
 Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); }
 double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }
 double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }
 double Length(Vector A) { return sqrt(Dot(A, A)); }
 Vector Normal(Vector A) { double l = Length(A); return Vector(-A.y/l, A.x/l); }

 double PolygonArea(const vector<Point>& p)
 {
     double ans = 0.0;
     int n = p.size();
     for(int i = ; i < n-; ++i)
         ans += Cross(p[i]-p[], p[i+]-p[]);
     return ans/;
 }

 struct Line
 {
     Point p;
     Vector v;
     double ang;
     Line() {}
     Line(Point p, Vector v):p(p), v(v) { ang = atan2(v.y, v.x); }
     bool operator < (const Line& L) const
     {
         return ang < L.ang;
     }
 };

 bool OnLeft(const Line& L, const Point& p)
 { return Cross(L.v, p-L.p) > ; }

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

 vector<Point> HalfplaneIntersection(vector<Line> L)
 {
     int n = L.size();
     sort(L.begin(), L.end());

     int first, last;
     vector<Point> p(n);
     vector<Line> q(n);
     vector<Point> ans;

     q[first=last=] = L[];
     for(int i = ; i < n; ++i)
     {
         while(first < last && !OnLeft(L[i], p[last-])) last--;
         while(first < last && !OnLeft(L[i], p[first])) first++;
         q[++last] = L[i];
         if(fabs(Cross(q[last].v, q[last-].v)) < eps)    //Èç¹ûÁ½Ö±ÏßƽÐУ¬È¡ÄÚ²àµÄÄǸö
         {
             last--;
             if(OnLeft(q[last], L[i].p)) q[last] = L[i];
         }
         if(first < last) p[last-] = GetLineIntersection(q[last-], q[last]);
     }
     while(first < last && !OnLeft(q[first], p[last-])) last--;
     if(last - first <= )    return ans;
     p[last] = GetLineIntersection(q[last], q[first]);

     for(int i = first; i <= last; ++i)    ans.push_back(p[i]);
     return ans;
 }

 int main(void)
 {
     #ifdef LOCAL
         freopen("3890in.txt", "r", stdin);
     #endif

     int n;
     while(scanf("%d", &n) ==  && n)
     {
         vector<Point> p, v, normal;
         int m, x, y;
         for(int i = ; i < n; ++i) { scanf("%d%d", &x, &y); p.push_back(Point(x, y)); }
         if(PolygonArea(p) < ) reverse(p.begin(), p.end());

         for(int i = ; i < n; ++i)
         {
             v.push_back(p[(i+)%n] - p[i]);
             normal.push_back(Normal(v[i]));
         }

         double left = , right = ;
         while(right - left > 5e-)
         {
             vector<Line> L;
             double mid = (right + left) / ;
             for(int i = ; i < n; ++i) L.push_back(Line(p[i] + normal[i]*mid, v[i]));
             vector<Point> Poly = HalfplaneIntersection(L);
             if(Poly.empty())    right = mid;
             else left = mid;
         }
         printf("%.6lf\n", left);
     }

     return ;
 }

代码君