uva 12818 Arc and Point && csu1503: 点到圆弧的距离

本文提供了一个计算几何问题的解决方案，通过求解三角形内圆心并判断点位置关系，使用C++实现算法流程，包括计算点到点的距离、确定圆心坐标及角度比较等关键步骤。

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此题先求出圆心，然后判断p点是否与b点在同一边，只要考虑两种情况就行了

<pre name="code" class="cpp">#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
struct point {
	double x,y;
	point() {}
	point(double _x,double _y) {
		x=_x;
		y=_y;
	}


} a,b,c,p;
point wx(point a,point b,point c) {
	double a1=b.x-a.x,b1=b.y-a.y,c1=(a1*a1+b1*b1)/2;
	double a2=c.x-a.x,b2=c.y-a.y,c2=(a2*a2+b2*b2)/2;
	double d=a1*b2-a2*b1;
	return point(a.x+(c1*b2-c2*b1)/d, a.y+(a1*c2-a2*c1)/d);
}
double ds(point a,point b) {
	double ans=sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
	return ans;
}
double min(double a,double b) {
	double ans=a<b?a:b;
	return ans;

}
double max(double a,double b) {
	return a>b?a:b;
}
double cha(point a,point b) {
	return (a.x*b.y-a.y*b.x);
}
int main() {
	int t=1;
	point xin;
	while(scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y,&c.x,&c.y,&p.x,&p.y)!=EOF) {
		double x,y,z,ans=0,tmp=0;
		xin=wx(a,b,c);
		x=max(ds(xin,a)-ds(xin,p),ds(xin,p)-ds(xin,a));
		y=(ds(p,a));
		z=(ds(p,c));
		double f1=atan2(a.y-xin.y,a.x-xin.x);
		double f2=atan2(b.y-xin.y,b.x-xin.x);
		double f3=atan2(c.y-xin.y,c.x-xin.x);
		double f4=atan2(p.y-xin.y,p.x-xin.x);
		printf("Case %d: ",t++);
		if(f1<f3) {
			if((f1<=f2&&f3>=f2)==(f1<=f4&&f3>=f4)) {
				printf("%.3lf\n",x);
				continue;
			} else {
				printf("%.3lf\n",min(y,z));
				continue;
			}

		} else {
			if((f3<=f2&&f1>=f2)==(f3<=f4&&f1>=f4)) {
				printf("%.3lf\n",x);
				continue;
			} else {
				printf("%.3lf\n",min(y,z));
				continue;
			}

		}


	}
}