就是……判断p与圆心的连线与圆的交点在不在圆弧上，在就是它到p的距离，不在就是p跟端点的最短距离

#include<iostream>
#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
const double pi=acos(-1.0);
struct dot
{
	double x,y;
	dot(){}
	dot(double a,double b){x=a;y=b;}
	dot operator +(dot a){return dot(x+a.x,y+a.y);}
	dot operator -(dot a){return dot(x-a.x,y-a.y);}
	dot operator *(double a){return dot(x*a,y*a);}
	double operator *(dot a){return x*a.y-y*a.x;}
	dot operator /(double a){return dot(x/a,y/a);}
	double operator /(dot a){return x*a.x+y*a.y;}
	bool operator ==(dot a){return x==a.x&&y==a.y;}
	dot norv(){return dot(-y,x);}
	dot univ(){double a=mod();return dot(x/a,y/a);}
	dot ro(double a){return dot(x*cos(a)-y*sin(a),x*sin(a)+y*cos(a));}
	double mod(){return sqrt(x*x+y*y);}
	double dis(dot a){return sqrt(pow(x-a.x,2)+pow(y-a.y,2));}
	friend istream & operator >> (istream &is,dot &a)
	{
		is>>a.x>>a.y;
		return is;
	}
	friend ostream & operator << (ostream &os,dot a)
	{
		os<<a.x<<" "<<a.y;
		return os;
	}
};
dot o,a,b,c,p;
double r;
void cg()
{
	dot A,B,C;
	A=a-o;
	B=b-o;
	C=c-o;
	if(A*B<0)
	{
		if(A*C<0&&B*C>0)
			return;
	}
	else
	{
		if(!(A*C>0&&B*C<0))
			return;
	}
	swap(a,c);
}
double work()
{
	dot A,B,C;
	A=a-o;
	B=p-o;
	C=c-o;
	if(A*B<0)
	{
		if(A*C<0&&B*C>0)
			return fabs(r-p.dis(o));
	}
	else
	{
		if(!(A*C>0&&B*C<0))
			return fabs(r-p.dis(o));
	}
	return min(p.dis(a),p.dis(c));
}
void cor()
{
	double A,B,C,D,E,F,G,H,I;
	A=2*(a.x-b.x);
	B=2*(a.y-b.y);
	C=pow(a.x,2)-pow(b.x,2)+pow(a.y,2)-pow(b.y,2);
	D=2*(a.x-c.x);
	E=2*(a.y-c.y);
	F=pow(a.x,2)-pow(c.x,2)+pow(a.y,2)-pow(c.y,2);
	G=dot(A,D)*dot(B,E);
	H=dot(C,F)*dot(B,E);
	I=dot(A,D)*dot(C,F);
	o=dot(H/G,I/G);
	r=o.dis(a);
}
int main()
{
	int jishu=0;
	double ans;
	while(cin>>a>>b>>c>>p)
	{
		cor();
		cg();
		ans=work();
		printf("Case %d: %.3f\n",++jishu,ans);
	}
}

csu 1503: 点到圆弧的距离-湖南省第十届大学生计算机程序设计竞赛