uva 11168 - Airport
凸包+一点直线的知识；
1 #include <cstdio> 2 #include <cmath> 3 #include <cstring> 4 #include <algorithm> 5 #include <vector> 6 #define eps 1e-9 7 using namespace std; 8 const double pi = acos(-1); 9 10 int dcmp(double x) 11 { 12 return fabs(x) < eps ? 0 : (x > 0 ? 1 : -1); 13 } 14 15 struct Point 16 { 17 double x; 18 double y; 19 20 Point(double x = 0, double y = 0):x(x), y(y) {} 21 22 bool operator < (const Point& e) const 23 { 24 return dcmp(x - e.x) < 0 || (dcmp(x - e.x) == 0 && dcmp(y - e.y) < 0); 25 } 26 27 bool operator == (const Point& e) const 28 { 29 return dcmp(x - e.x) == 0 && dcmp(y - e.y) == 0; 30 } 31 }; 32 33 typedef Point Vector; 34 35 Vector operator + (Point A, Point B) 36 { 37 return Vector(A.x + B.x, A.y + B.y); 38 } 39 40 Vector operator - (Point A, Point B) 41 { 42 return Vector(A.x - B.x, A.y - B.y); 43 } 44 45 Vector operator * (Point A, double p) 46 { 47 return Vector(A.x * p, A.y * p); 48 } 49 50 Vector operator / (Point A, double p) 51 { 52 return Vector(A.x / p, A.y / p); 53 } 54 55 double cross(Point a,Point b){return a.x*b.y-a.y*b.x;} 56 Point rotate(Point a,double ang){return Point(a.x*cos(ang)-a.y*sin(ang),a.x*sin(ang)+a.y*cos(ang));} 57 int convexhull(Point *p,int n,Point *ch) 58 { 59 sort(p,p+n); 60 int m=0; 61 for(int i=0;i<n;i++) 62 { 63 while(m>1&&cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)m--; 64 ch[m++]=p[i]; 65 } 66 int k=m; 67 for(int i=n-2;i>=0;i--) 68 { 69 while(m>k&&cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)m--; 70 ch[m++]=p[i]; 71 } 72 if(n>1)m--; 73 return m; 74 } 75 Point p[100009],ch[100009]; 76 77 int main() 78 { 79 int t,n,ca=1; 80 scanf("%d",&t); 81 while(t--) 82 { 83 double sumx=0; 84 double sumy=0; 85 scanf("%d",&n); 86 for(int i=0;i<n;i++) 87 { 88 scanf("%lf%lf",&p[i].x,&p[i].y); 89 sumx+=p[i].x; 90 sumy+=p[i].y; 91 } 92 printf("Case #%d: ",ca++); 93 double mi=99999999999.0; 94 int m=convexhull(p,n,ch); 95 for(int i=0;i<m;i++) 96 { 97 double a=-ch[(i+1)%m].y+ch[i].y; 98 double b=ch[(i+1)%m].x-ch[i].x; 99 double c=ch[i].x*ch[(i+1)%m].y-ch[i].y*ch[(i+1)%m].x; 100 double d=fabs(a*sumx+b*sumy+c*n)/sqrt(a*a+b*b); 101 mi=min(mi,d); 102 } 103 printf("%.3lf\n",n>2?mi/n:0); 104 } 105 return 0; 106 }
