LA 3890 半平面交

二分查询答案，判断每一个新形成的向量合在一块能否形成半平面交

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 using namespace std;
 #define N 110
 #define eps 1e-7

 int dcmp(double x)
 {
     if(fabs(x)<eps) return ;
     else return x<?-:;
 }

 struct Point{
     double x , y;
     Point(double x= , double y=):x(x),y(y){}
 }po[N] , poly[N];

 typedef Point Vector;
 Vector vec[N]; //记录每条边上对应的法向量

 Vector operator+(Vector a , Vector b) { return Vector(a.x+b.x , a.y+b.y); }
 Vector operator-(Point a , Point b) { return Vector(a.x-b.x , a.y-b.y); }
 Vector operator*(Vector a , double b) { return Vector(a.x*b , a.y*b); }
 Vector operator/(Vector a , double b) { return Vector(a.x/b , a.y/b); }
 bool operator==(const Point &a , const Point &b) { return dcmp(a.x-b.x) ==  && dcmp(a.y-b.y) == ; }

 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 - b.x*a.y; }
 double Length(Vector a) { return sqrt(Dot(a,a)); }
 double Angle(Vector A , Vector B) { return acos(Dot(A,B)) / Length(A) / Length(B); }
 double Area2(Point A , Point B , Point C) { return Cross(B-A , C-A); }

 Vector Rotate(Vector A , double rad) { return Vector(A.x*cos(rad)-A.y*sin(rad) , A.x*sin(rad)+A.y*cos(rad)); }

 Vector Normal(Vector a)
 {
     double l = Length(a);
     return Vector(-a.y/l , a.x/l);
 }

 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 &m)const {
         return ang<m.ang;
     }
 }line[N] , L[N];

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

 Point GetIntersection(Line a , 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;
 }

 /***半平面交的主过程，返回形成半平面交点的个数，无法形成就返回0***/
 int HalfplaneIntersection(Line *L , int n , Point *poly)
 {
     sort(L , L+n);
     int first , last;
     Point *p = new Point[n];
     Line *q = new Line[n];
     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)
         {
             last--;
             if(OnLeft(q[last] , L[i].P)) q[last]=L[i];
         }
         if(first<last) p[last-] = GetIntersection(q[last-] , q[last]);
     }
     while(first < last && !OnLeft(q[first] , p[last-])) last--;
     //删除无用平面
     if(last-first<=) return ;
     p[last] = GetIntersection(q[last] , q[first]);

     //从deque复制到输出中
     int m=;
     for(int i=first ; i<=last ; i++) poly[m++] = p[i];
     return m;
 }

 int main()
 {
    // freopen("a.in" , "r" , stdin);
     int n;
     while(scanf("%d" , &n) , n)
     {
         for(int i= ; i<n ; i++)
             scanf("%lf%lf" , &po[i].x , &po[i].y);

         for(int i= ; i<n ; i++) vec[i] = Normal(po[(i+)%n]-po[i]);
         double l= , r=;
         while(r-l>eps){
             double m=(l+r)/;
             for(int i= ; i<n ; i++) L[i] = Line(po[i]+vec[i]*m , po[(i+)%n]-po[i]);
             if(HalfplaneIntersection(L , n , poly)>) l=m;
             else r=m;
         }
         printf("%.6f\n" , l);
     }
     return ;
 }