CSU 1503 点到圆弧的距离（2014湖南省程序设计竞赛A题）

  1 #include<cstdio>
  2 #include<cstring>
  3 #include<iostream>
  4 #include<algorithm>
  5 #include<cmath>
  6 using namespace std;
  7 const double eps = 1e-6,PI = acos(-1.0);
  8 
  9 struct Point
 10 {
 11     double x,y;
 12     friend Point operator - (Point a,Point b)
 13     {
 14         Point temp;
 15         temp.x = a.x - b.x;
 16         temp.y = a.y - b.y;
 17         return temp;
 18     }
 19 };
 20 
 21 Point p1,p2,p3,pc,pp;
 22 double r;
 23 
 24 Point get_pc1(Point p1, Point p2, Point p3)    //求圆心
 25 {
 26   Point p;
 27   if(fabs(p1.y-p2.y) > eps)      //因为计算过程中有出现(y2-y1)做分母的情况，所以这里分了两种情况讨论
 28   {
 29  //      double t1 = p3.x*p3.x - p1.x*p1.x+p3.y*p3.y-p1.y*p1.y-((p3.y-p1.y)*(p1.x*p1.x-p2.x*p2.x+p1.y*p1.y-p2.y*p2.y)) / (p2.y-p1.y);
 30    //    double t2 = 2.0*(p3.x-p1.x)-(2*(p3.y-p1.y)*(p2.x-p1.x)) / (p2.y-p1.y);
 31         double  t1 = (p2.x*p2.x-p1.x*p1.x+p2.y*p2.y-p1.y*p1.y) *(p3.y-p1.y) - (p2.y-p1.y)*(p3.x*p3.x-p1.x*p1.x+p3.y*p3.y-p1.y*p1.y);
 32         double t2 = 2.0*((p3.y-p1.y)*(p2.x-p1.x) - (p2.y-p1.y)*(p3.x-p1.x));
 33        p.x = t1 / t2;
 34        p.y = (p2.x*p2.x-p1.x*p1.x+p2.y*p2.y-p1.y*p1.y-2*(p2.x-p1.x)*p.x) / (2*(p2.y-p1.y));
 35     }
 36     else
 37     {
 38         p.x = (p2.x*p2.x-p1.x*p1.x+p2.y*p2.y-p1.y*p1.y) / (2*(p2.x-p1.x));
 39         p.y = (p3.x*p3.x-p1.x*p1.x+p3.y*p3.y-p1.y*p1.y-2*(p3.x-p1.x)*p.x) / (2*(p3.y-p1.y));
 40     }
 41     return p;
 42 }
 43 double dis(Point a,Point b)
 44 {
 45     return sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y));
 46 }
 47 double mult_ca(Point p1,Point p2)   //叉乘
 48 {
 49     return p1.x * p2.y - p2.x * p1.y;
 50 }
 51 
 52 double get_ans(Point pc,Point pp,Point p1,Point p2,Point p3)
 53 {
 54     double temp = mult_ca(p2-p1,p3-p1);
 55     double f1 = atan2((p1-pc).x,(p1-pc).y);
 56     double f2 = atan2((p3-pc).x,(p3-pc).y);
 57     double f3 = atan2((p2-pc).x,(p2-pc).y);
 58     double f4 = atan2((pp-pc).x,(pp-pc).y);
 59     double ans1 = fabs(dis(pp,pc)-dis(p1,pc));
 60     double ans2 = min(dis(pp,p1),dis(pp,p3));
 61     if(temp < 0)    //顺时针给点
 62     {
 63         if(f1 < f2)      //判断区间方向，这样有利于判断p点和p2点是不是在同一个区间
 64         {
 65             if((f3 >= f1 && f3 <= f2) == (f4 >= f1 && f4 <= f2) ) return ans1;
 66             else return ans2;
 67         }
 68         else
 69         {
 70             if((f3 >= f2 && f3 <= f1) == (f4 >=f2 && f4 <= f1) ) return ans1;
 71             else return ans2;
 72         }
 73     }
 74     else
 75     {
 76         if(f2 < f1)
 77         {
 78             if((f3 >= f2 && f3 <= f1) == (f4 >= f2 && f4 <= f1) ) return ans1;
 79             else return ans2;
 80         }
 81         else
 82         {
 83             if((f3 >= f1 && f3 <= f2) == (f4 >= f1 && f4 <= f2)) return ans1;
 84             else return ans2;
 85         }
 86     }
 87 }
 88 
 89 int main()
 90 {
 91   //   freopen("in","r",stdin);
 92     int kase = 1;
 93     while(scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&p1.x,&p1.y,&p2.x,&p2.y,&p3.x,&p3.y,&pp.x,&pp.y)!=EOF)
 94     {
 95         pc = get_pc1(p1,p2,p3);
 96         double ans = get_ans(pc,pp,p1,p2,p3);
 97         printf("Case %d: %.3lf\n",kase++,ans);
 98     }
 99     return 0;
100 }