51nod 1298 圆与三角形 (计算几何)

http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1298
求出圆心到三条线段的最短距离,然后判断是否有顶点在圆外,就把全部情况举出来。
  1 #include <cstdio>
  2 #include <cstring>
  3 #include <cmath>
  4 #include <algorithm>
  5 using namespace std;
  6 const double PI = acos(-1.0);
  7 double torad(double deg) { return deg/180 * PI; }
  8 
  9 struct Point
 10 {
 11   double x, y;
 12   Point(double x=0, double y=0):x(x),y(y) { }
 13 };
 14 
 15 typedef Point Vector;
 16 
 17 Vector operator + (const Vector& A, const Vector& B) { return Vector(A.x+B.x, A.y+B.y); }
 18 Vector operator - (const Point& A, const Point& B) { return Vector(A.x-B.x, A.y-B.y); }
 19 Vector operator * (const Vector& A, double p) { return Vector(A.x*p, A.y*p); }
 20 Vector operator / (const Vector& A, double p) { return Vector(A.x/p, A.y/p); }
 21 
 22 bool operator < (const Point& a, const Point& b)    //结构体运算符的重载
 23 {
 24   return a.x < b.x || (a.x == b.x && a.y < b.y);
 25 }
 26 
 27 const double eps = 1e-8;
 28 int dcmp(double x) { if(fabs(x) < eps) return 0; else return x < 0 ? -1 : 1; }
 29 
 30 bool operator == (const Point& a, const Point &b)
 31 {
 32   return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0;
 33 }
 34 
 35 //基本运算：
 36 double dist(const Vector& A, const Vector& B) {return sqrt(pow(A.x-B.x,2)+pow(A.y-B.y,2));}
 37 double Dot(const Vector& A, const Vector& B) { return A.x*B.x + A.y*B.y; }
 38 double Length(const Vector& A) { return sqrt(Dot(A, A)); }
 39 double Angle(const Vector& A, const Vector& B) { return acos(Dot(A, B) / Length(A) / Length(B)); }
 40 double Cross(const Vector& A, const Vector& B) { return A.x*B.y - A.y*B.x; }
 41 double Area2(Point A, Point B, Point C) {return Cross(B-A, C-A);}
 42 
 43 //向量旋转 rad是弧度
 44 Vector Rotate(const Vector& A, double rad)
 45 {
 46   return Vector(A.x*cos(rad)-A.y*sin(rad), A.x*sin(rad)+A.y*cos(rad));
 47 }
 48 //点和直线：
 49 //两直线的交点
 50 Point GetLineIntersection(const Point& P, const Point& v, const Point& Q, const Point& w)
 51 {
 52   Vector u = P-Q;
 53   double t = Cross(w, u) / Cross(v, w);
 54   return P+v*t;
 55 }
 56 
 57 //点到直线的距离
 58 double DistanceToLine(const Point& P, const Point& A, const Point& B)
 59 {
 60   Vector v1=B-A, v2=P-A;
 61   return fabs(Cross(v1,v2)) / Length(v1);
 62 }
 63 
 64 //点到线段的距离
 65 double DistanceToSegment(const Point& P, const Point& A, const Point& B)
 66 {
 67   if(A == B) return Length(P-A);
 68   Vector v1 = B - A, v2 = P - A, v3 = P - B;
 69   if(dcmp(Dot(v1, v2)) < 0) return Length(v2);
 70   else if(dcmp(Dot(v1, v3)) > 0) return Length(v3);
 71   else return fabs(Cross(v1, v2)) / Length(v1);
 72 }
 73 
 74 //点在直线上的投影
 75 Point GetLineProjection(const Point &P, const Point &A,const Point &B)
 76 {
 77     Vector v = B - A;
 78     return A+v*(Dot(v, P-A) / Dot(v, v));
 79 }
 80 
 81 //线段相交判定
 82 bool SegmentProperIntersection(const Point& a1, const Point& a2, const Point& b1, const Point& b2)
 83 {
 84   double c1 = Cross(a2-a1,b1-a1), c2 = Cross(a2-a1,b2-a1),
 85   c3 = Cross(b2-b1,a1-b1), c4=Cross(b2-b1,a2-b1);
 86   return dcmp(c1)*dcmp(c2)<0 && dcmp(c3)*dcmp(c4)<0;
 87 }
 88 
 89 
 90 //判断点在线段上(两个端点除外)
 91 bool OnSegment(const Point& p, const Point& a1, const Point& a2)
 92 {
 93   return dcmp(Cross(a1-p, a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) < 0;
 94 }
 95 
 96 int main()
 97 {
 98     freopen("a.txt","r",stdin);
 99     int t;
100     Vector s,p[5];
101     double r,x;
102     scanf("%d",&t);
103     while(t--)
104     {
105         scanf("%lf%lf%lf",&s.x,&s.y,&r);
106         for(int i=0;i<3;i++)
107             scanf("%lf%lf",&p[i].x,&p[i].y);
108         int ans=0;
109         x=DistanceToSegment(s,p[0],p[1]);
110         if(x<=r) ans++;
111         x=DistanceToSegment(s,p[0],p[2]);
112         if(x<=r) ans++;
113         x=DistanceToSegment(s,p[1],p[2]);
114         if(x<=r) ans++;
115         bool flag=0;
116         for(int i=0;i<3;i++)
117         {
118             if(dist(s,p[i])>r)
119             {
120                 flag=1;break;
121             }
122         }
123      //   printf("%d %d
",ans,flag);
124         if(ans>=1&&flag) puts("Yes");
125         else puts("No");
126     }
127     return 0;
128 }