uva 1396 - Most Distant Point from the Sea

半平面的交，二分的方法；

  1 #include<cstdio>
  2 #include<algorithm>
  3 #include<cmath>
  4 #define eps 1e-6
  5 using namespace std;
  6 
  7 int dcmp(double x)
  8 {
  9     return fabs(x) < eps ? 0 : (x > 0 ? 1 : -1);
 10 }
 11 
 12 struct Point
 13 {
 14     double x;
 15     double y;
 16     Point(double x = 0, double y = 0):x(x), y(y) {}
 17 };
 18 typedef Point Vector;
 19 
 20 Vector operator + (Point A, Point B)
 21 {
 22     return Vector(A.x + B.x, A.y + B.y);
 23 }
 24 
 25 Vector operator - (Point A, Point B)
 26 {
 27     return Vector(A.x - B.x, A.y - B.y);
 28 }
 29 
 30 Vector operator * (Point A, double p)
 31 {
 32     return Vector(A.x * p, A.y * p);
 33 }
 34 
 35 Vector operator / (Point A, double p)
 36 {
 37     return Vector(A.x / p, A.y / p);
 38 }
 39 double dot(Point a,Point b)
 40 {
 41     return a.x*b.x+a.y*b.y;
 42 }
 43 double cross(Point a,Point b)
 44 {
 45     return a.x*b.y-a.y*b.x;
 46 }
 47 
 48 Vector nomal(Vector a)
 49 {
 50     double l=sqrt(dot(a,a));
 51     return Vector(-a.y/l,a.x/l);
 52 }
 53 
 54 struct line
 55 {
 56     Point p;
 57     Vector v;
 58     double ang;
 59     line() {}
 60     line(Point p,Vector v):p(p),v(v)
 61     {
 62         ang=atan2(v.y,v.x);
 63     }
 64     bool operator<(const line &t)const
 65     {
 66         return ang<t.ang;
 67     }
 68 };
 69 
 70 bool onleft(line l,Point p)
 71 {
 72     return (cross(l.v,p-l.p)>0);
 73 }
 74 
 75 Point getintersection(line a,line b)
 76 {
 77     Vector u=a.p-b.p;
 78     double t=cross(b.v,u)/cross(a.v,b.v);
 79     return a.p+a.v*t;
 80 }
 81 
 82 int halfplanintersection(line *l,int n,Point *poly)
 83 {
 84     sort(l,l+n);
 85     int first,last;
 86     Point *p=new Point[n];
 87     line *q=new line[n];
 88     q[first=last=0]=l[0];
 89     for(int i=1; i<n; i++)
 90     {
 91         while(first<last && !onleft(l[i],p[last-1]))last--;
 92         while(first<last && !onleft(l[i],p[first]))first++;
 93         q[++last]=l[i];
 94         if(fabs(cross(q[last].v,q[last-1].v))<eps)
 95         {
 96             last--;
 97             if(onleft(q[last],l[i].p))q[last]=l[i];
 98         }
 99         if(first<last)p[last-1]=getintersection(q[last-1],q[last]);
100     }
101     while(first<last && !onleft(q[first],p[last-1]))last--;
102     if((last-first )<=1)return 0;
103     p[last]=getintersection(q[last],q[first]);
104     int m=0;
105     for(int i=first; i<=last; i++)poly[m++]=p[i];
106     return m;
107 }
108 
109 Point p[200],poly[200];
110 line l[200];
111 Point v[200],v2[200];
112 int main()
113 {
114     int n,m;
115     double x,y;
116     while(scanf("%d",&n)&&n)
117     {
118         for(int i=0; i<n; i++)
119         {
120             scanf("%lf%lf",&x,&y);
121             p[i]=Point(x,y);
122         }
123         for(int i=0; i<n; i++)
124         {
125             v[i]=p[(i+1)%n]-p[i];
126             v2[i]=nomal(v[i]);
127         }
128         double left=0.0,right=20000.0;
129         while(right-left>eps)
130         {
131             double mid=(left+right)/2;
132             for(int i=0; i<n; i++)
133                 l[i]=line(p[i]+v2[i]*mid,v[i]);
134             m=halfplanintersection(l,n,poly);
135             if(!m)right=mid;
136             else left=mid;
137         }
138         printf("%.6lf\n",left);
139     }
140     return 0;
141 }
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posted @ 2013-11-05 13:13  Yours1103  阅读(173)  评论(0)  编辑  收藏  举报