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

  1 #include <iostream>

  2 #include <cstdio>

  3 #include <cstring>

  4 #include <cmath>

  5 #include <algorithm>

  6 using namespace std;

  7 #define N 110

  8 #define eps 1e-7

  9 

 10 int dcmp(double x)

 11 {

 12     if(fabs(x)<eps) return 0;

 13     else return x<0?-1:1;

 14 }

 15 

 16 struct Point{

 17     double x , y;

 18     Point(double x=0 , double y=0):x(x),y(y){}

 19 }po[N] , poly[N];

 20 

 21 typedef Point Vector;

 22 Vector vec[N]; //记录每条边上对应的法向量

 23 

 24 Vector operator+(Vector a , Vector b) { return Vector(a.x+b.x , a.y+b.y); }

 25 Vector operator-(Point a , Point b) { return Vector(a.x-b.x , a.y-b.y); }

 26 Vector operator*(Vector a , double b) { return Vector(a.x*b , a.y*b); }

 27 Vector operator/(Vector a , double b) { return Vector(a.x/b , a.y/b); }

 28 bool operator==(const Point &a , const Point &b) { return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0; }

 29 

 30 double Dot(Vector a  , Vector b) { return a.x*b.x+a.y*b.y; }

 31 double Cross(Vector a , Vector b) { return a.x*b.y - b.x*a.y; }

 32 double Length(Vector a) { return sqrt(Dot(a,a)); }

 33 double Angle(Vector A , Vector B) { return acos(Dot(A,B)) / Length(A) / Length(B); }

 34 double Area2(Point A , Point B , Point C) { return Cross(B-A , C-A); }

 35 

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

 37 

 38 Vector Normal(Vector a)

 39 {

 40     double l = Length(a);

 41     return Vector(-a.y/l , a.x/l);

 42 }

 43 

 44 struct Line{

 45     Point P;

 46     Vector v;

 47     double ang;

 48     Line(){}

 49     Line(Point P , Vector v):P(P),v(v){ang = atan2(v.y,v.x);}

 50     bool operator<(const Line &m)const {

 51         return ang<m.ang;

 52     }

 53 }line[N] , L[N];

 54 

 55 bool OnLeft(Line L , Point P)

 56 {

 57     return Cross(L.v , P-L.P) > 0;

 58 }

 59 

 60 Point GetIntersection(Line a , Line b)

 61 {

 62     Vector u = a.P-b.P;

 63     double t = Cross(b.v , u) / Cross(a.v , b.v);

 64     return a.P+a.v*t;

 65 }

 66 

 67 /***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***/

 68 int HalfplaneIntersection(Line *L , int n , Point *poly)

 69 {

 70     sort(L , L+n);

 71     int first , last;

 72     Point *p = new Point[n];

 73     Line *q = new Line[n];

 74     q[first=last=0] = L[0];

 75     for(int i=1 ; i<n ; i++)

 76     {

 77         while(first<last && !OnLeft(L[i] , p[last-1])) last--;

 78         while(first<last && !OnLeft(L[i] , p[first])) first++;

 79         q[++last] = L[i];

 80         if(fabs(Cross(q[last].v , q[last-1].v)) < eps)

 81         {

 82             last--;

 83             if(OnLeft(q[last] , L[i].P)) q[last]=L[i];

 84         }

 85         if(first<last) p[last-1] = GetIntersection(q[last-1] , q[last]);

 86     }

 87     while(first < last && !OnLeft(q[first] , p[last-1])) last--;

 88     //删除无用平面

 89     if(last-first<=1) return 0;

 90     p[last] = GetIntersection(q[last] , q[first]);

 91 

 92     //从deque复制到输出中

 93     int m=0;

 94     for(int i=first ; i<=last ; i++) poly[m++] = p[i];

 95     return m;

 96 }

 97 

 98 int main()

 99 {

100    // freopen("a.in" , "r" , stdin);

101     int n;

102     while(scanf("%d" , &n) , n)

103     {

104         for(int i=0 ; i<n ; i++)

105             scanf("%lf%lf" , &po[i].x , &po[i].y);

106 

107         for(int i=0 ; i<n ; i++) vec[i] = Normal(po[(i+1)%n]-po[i]);

108         double l=0 , r=20000;

109         while(r-l>eps){

110             double m=(l+r)/2;

111             for(int i=0 ; i<n ; i++) L[i] = Line(po[i]+vec[i]*m , po[(i+1)%n]-po[i]);

112             if(HalfplaneIntersection(L , n , poly)>0) l=m;

113             else r=m;

114         }

115         printf("%.6f\n" , l);

116     }

117     return 0;

118 }