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。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const double PI = acos(-1.0);
double torad(double deg) { return deg/180 * PI; }

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

typedef Point Vector;

Vector operator + (const Vector& A, const Vector& B) { return Vector(A.x+B.x, A.y+B.y); }
Vector operator - (const Point& A, const Point& B) { return Vector(A.x-B.x, A.y-B.y); }
Vector operator * (const Vector& A, double p) { return Vector(A.x*p, A.y*p); }
Vector operator / (const Vector& A, double p) { return Vector(A.x/p, A.y/p); }

bool operator < (const Point& a, const Point& b)    //结构体运算符的重载
{
  return a.x < b.x || (a.x == b.x && a.y < b.y);
}

const double eps = 1e-8;
int dcmp(double x) { if(fabs(x) < eps) return 0; else return x < 0 ? -1 : 1; }

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

//基本运算：
double dist(const Vector& A, const Vector& B) {return sqrt(pow(A.x-B.x,2)+pow(A.y-B.y,2));}
double Dot(const Vector& A, const Vector& B) { return A.x*B.x + A.y*B.y; }
double Length(const Vector& A) { return sqrt(Dot(A, A)); }
double Angle(const Vector& A, const Vector& B) { return acos(Dot(A, B) / Length(A) / Length(B)); }
double Cross(const Vector& A, const Vector& B) { return A.x*B.y - A.y*B.x; }
double Area2(Point A, Point B, Point C) {return Cross(B-A, C-A);}

//向量旋转 rad是弧度
Vector Rotate(const Vector& A, double rad)
{
  return Vector(A.x*cos(rad)-A.y*sin(rad), A.x*sin(rad)+A.y*cos(rad));
}
//点和直线：
//两直线的交点
Point GetLineIntersection(const Point& P, const Point& v, const Point& Q, const Point& w)
{
  Vector u = P-Q;
  double t = Cross(w, u) / Cross(v, w);
  return P+v*t;
}

//点到直线的距离
double DistanceToLine(const Point& P, const Point& A, const Point& B)
{
  Vector v1=B-A, v2=P-A;
  return fabs(Cross(v1,v2)) / Length(v1);
}

//点到线段的距离
double DistanceToSegment(const Point& P, const Point& A, const Point& B)
{
  if(A == B) return Length(P-A);
  Vector v1 = B - A, v2 = P - A, v3 = P - B;
  if(dcmp(Dot(v1, v2)) < 0) return Length(v2);
  else if(dcmp(Dot(v1, v3)) > 0) return Length(v3);
  else return fabs(Cross(v1, v2)) / Length(v1);
}

//点在直线上的投影
Point GetLineProjection(const Point &P, const Point &A,const Point &B)
{
    Vector v = B - A;
    return A+v*(Dot(v, P-A) / Dot(v, v));
}

//线段相交判定
bool SegmentProperIntersection(const Point& a1, const Point& a2, const Point& b1, const Point& b2)
{
  double c1 = Cross(a2-a1,b1-a1), c2 = Cross(a2-a1,b2-a1),
  c3 = Cross(b2-b1,a1-b1), c4=Cross(b2-b1,a2-b1);
  return dcmp(c1)*dcmp(c2)<0 && dcmp(c3)*dcmp(c4)<0;
}


//判断点在线段上(两个端点除外)
bool OnSegment(const Point& p, const Point& a1, const Point& a2)
{
  return dcmp(Cross(a1-p, a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) < 0;
}

int main()
{
    freopen("a.txt","r",stdin);
    int t;
    Vector s,p[5];
    double r,x;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf%lf%lf",&s.x,&s.y,&r);
        for(int i=0;i<3;i++)
            scanf("%lf%lf",&p[i].x,&p[i].y);
        int ans=0;
        x=DistanceToSegment(s,p[0],p[1]);
        if(x<=r) ans++;
        x=DistanceToSegment(s,p[0],p[2]);
        if(x<=r) ans++;
        x=DistanceToSegment(s,p[1],p[2]);
        if(x<=r) ans++;
        bool flag=0;
        for(int i=0;i<3;i++)
        {
            if(dist(s,p[i])>r)
            {
                flag=1;break;
            }
        }
     //   printf("%d %d
",ans,flag);
        if(ans>=1&&flag) puts("Yes");
        else puts("No");
    }
    return 0;
}