题目描述
输入一个点 P 和一条圆弧（圆周的一部分），你的任务是计算 P 到圆弧的最短距离。换句话 说，你需要在圆弧上找一个点，到 P点的距离最小。
提示：请尽量使用精确算法。相比之下，近似算法更难通过本题的数据。
提示：请尽量使用精确算法。相比之下，近似算法更难通过本题的数据。
输入
输入包含最多 10000组数据。每组数据包含 8个整数 x1, y1, x2, y2, x3, y3, xp, yp。圆弧的起点 是 A(x1,y1)，经过点 B(x2,y2)，结束位置是 C(x3,y3)。点 P的位置是 (xp,yp)。输入保证 A, B, C 各不相同且不会共线。上述所有点的坐标绝对值不超过 20。
输出
对于每组数据，输出测试点编号和 P 到圆弧的距离，保留三位小数。
样例输入
0 0 1 1 2 0 1 -1
3 4 0 5 -3 4 0 1
样例输出
Case 1: 1.414 Case 2: 4.000
分两种情况：
第一种：点跟圆心的连线在那段扇形的圆弧范围内，点到圆弧的最短距离为点到圆心的距离减去半径然后取绝对值；
第二种：点跟圆心的连线不在那段扇形的圆弧范围内，点到圆弧的最短的距离为到这段圆弧的两个端点的最小值。
1 #include <bits/stdc++.h> 2 using namespace std; 3 const double eps = 1e-8; 4 const double PI = acos(-1.0); 5 int casee = 0; 6 int dcmp(double x) { 7 if(fabs(x) < eps) return 0; 8 else return x < 0 ? -1 : 1; 9 } 10 struct Vector { 11 double x, y; 12 Vector(double x_ = 0, double y_ = 0) : x(x_), y(y_) {} 13 14 Vector operator+ (const Vector &a) const { 15 return Vector(x + a.x, y + a.y); 16 } 17 Vector operator- (const Vector &a) const { 18 return Vector(x - a.x, y - a.y); 19 } 20 Vector operator* (double p) { 21 return Vector(x * p, y * p); 22 } 23 friend Vector operator* (double p, const Vector &a) { 24 return Vector(a.x * p, a.y * p); 25 } 26 Vector operator/ (double p) { 27 return Vector(x / p, y / p); 28 } 29 bool operator== (const Vector &a) const { 30 return dcmp(x - a.x) == 0 && dcmp(y - a.y) == 0; 31 } 32 33 friend double dmul(const Vector &a, const Vector &b) { 34 return a.x * b.x + a.y * b.y; 35 } 36 friend double cmul(const Vector &a, const Vector &b) { 37 return a.x * b.y - a.y * b.x; 38 } 39 friend double angle(const Vector &a, const Vector &b) { 40 return acos(dmul(a, b) / a.length() / b.length()); 41 } 42 friend Vector rotate(const Vector &a, double rad) { 43 return Vector(a.x * cos(rad) - a.y * sin(rad), a.x * sin(rad) + a.y * cos(rad)); 44 } 45 friend Vector normal(const Vector &a) { 46 double l = a.length(); 47 return Vector(a.x / l, a.y / l); 48 } 49 50 double length() const { 51 return sqrt(dmul(*this, *this)); 52 } 53 }; 54 typedef Vector Point; 55 double dis(Point A, Point B) { 56 return (A-B).length(); 57 } 58 Point pp1, pp2, pp3, pp, o; 59 60 Point get_cir(Point a,Point b,Point c) {//已知圆上3点,求圆的外心 61 double a1=b.x-a.x,b1=b.y-a.y,c1=(a1*a1+b1*b1)/2; 62 double a2=c.x-a.x,b2=c.y-a.y,c2=(a2*a2+b2*b2)/2; 63 double d=a1*b2-a2*b1; 64 return Point(a.x+(c1*b2-c2*b1)/d,a.y+(a1*c2-a2*c1)/d); 65 } 66 double AngleToX(Vector A, Vector B) {//向量与x轴正方向的夹角,范围[0,2*pi) B为(1,0) 67 double res = angle(A, B); 68 if(dcmp(cmul(A, B)) < 0) res = PI * 2 - res; 69 return res; 70 } 71 void gao (Point A,Point B,Point C,Point P,Point O) 72 { 73 Vector X(1, 0); 74 Vector OA = A - O, OP = P - O, OC = C - O, OB = B - O; 75 double fa = AngleToX(OA, X); 76 double fb = AngleToX(OB, X); 77 double fc = AngleToX(OC, X); 78 double fp = AngleToX(OP, X); 79 double ans = min(dis(P, A), dis(P, C)); 80 double r = dis(A, O); 81 if(fa>fc) swap(fa,fc); 82 if(dcmp(fa-fb)<=0 && dcmp(fb - fc) <= 0 && dcmp(fa - fp) <= 0 && dcmp(fp - fc) <= 0) { 83 double r = dis(A, O); 84 ans = min(ans, fabs(dis(O, P) - r)); 85 } 86 if(!(dcmp(fa - fb) <= 0 &&dcmp(fb-fc)<= 0)&&!(dcmp(fa-fp)<=0&&dcmp(fp-fc)<=0)) { 87 ans = min(ans, fabs(dis(O, P)- r)); 88 } 89 90 printf("Case %d: %.3f ", ++casee, ans); 91 } 92 int main() { 93 while(~scanf("%lf%lf%lf%lf%lf%lf%lf%lf", &pp1.x,&pp1.y,&pp2.x,&pp2.y,&pp3.x,&pp3.y,&pp.x,&pp.y)){ 94 o=get_cir(pp1,pp2,pp3); 95 gao(pp1,pp2,pp3,pp,o); 96 } 97 return 0; 98 }
解法2
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <cmath> 5 #include <map> 6 #define eps (1e-4) 7 #define N 205 8 #define dd double 9 #define sqr(x) ((x)*(x)) 10 const double pi = acos(-1); 11 using namespace std; 12 struct Point 13 { 14 double x,y; 15 }; 16 double cross(Point a,Point b,Point c) ///叉积 17 { 18 return (a.x-c.x)*(b.y-c.y)-(b.x-c.x)*(a.y-c.y); 19 } 20 double dis(Point a,Point b) ///距离 21 { 22 return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); 23 } 24 Point waixin(Point a,Point b,Point c) ///外接圆圆心坐标 25 { 26 Point p; 27 double a1 = b.x - a.x, b1 = b.y - a.y, c1 = (a1*a1 + b1*b1)/2; 28 double a2 = c.x - a.x, b2 = c.y - a.y, c2 = (a2*a2 + b2*b2)/2; 29 double d = a1*b2 - a2*b1; 30 p.x = a.x + (c1*b2 - c2*b1)/d, p.y=a.y + (a1*c2 -a2*c1)/d; 31 return p; 32 } 33 bool judge(Point a,Point b,Point c,Point p){ ///此模板判断点 p 是否在两条射线 ab 和 ac 之间(但是p点不在 ab或者 ac上) 34 if(cross(b,c,a)>0&&cross(p,a,c)<0&&cross(p,a,b)>0){ ///b 在 c 的逆时针方向 35 return true; 36 } 37 if(cross(b,c,a)<0&&cross(p,a,c)>0&&cross(p,a,b)<0){ ///b 在 c 的顺时针方向 38 return true; 39 } 40 return false; 41 } 42 bool judge1(Point a,Point b,Point c,Point p){ ///此模板判断点 p 是否在两条射线 ab 和 ac 之间(p点可以在 ab或者 ac上) 43 if(cross(b,c,a)>0&&cross(p,a,c)<=0&&cross(p,a,b)>=0){ ///b 在 c 的逆时针方向 44 return true; 45 } 46 if(cross(b,c,a)<0&&cross(p,a,c)>=0&&cross(p,a,b)<=0){ ///b 在 c 的顺时针方向 47 return true; 48 } 49 return false; 50 } 51 int main() 52 { 53 Point a,b,c,p; 54 int t = 1; 55 freopen("a.in","r",stdin); 56 freopen("a.txt","w",stdout); 57 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) 58 { 59 60 Point circle = waixin(a,b,c); 61 double r = dis(circle,a); 62 double ans = min(dis(p,a),dis(p,c)); 63 double op = dis(circle,p); 64 if(fabs(2*r-dis(a,c))<eps){ ///平角特殊处理 65 if(cross(p,c,circle)*cross(b,c,circle)>=0){ 66 ans = min(ans,fabs(op-r)); 67 } 68 printf("Case %d: %.3lf ",t++,ans); 69 continue; 70 } 71 if(judge(circle,a,c,b)) ///劣弧 72 { 73 if(judge1(circle,a,c,p)) ans = min(ans,fabs(op-r)); 74 } 75 else 76 { 77 if(!judge(circle,a,c,p)) ans = min(ans,fabs(op-r)); 78 } 79 printf("Case %d: %.3lf ",t++,ans); 80 } 81 }
标程:
1 // Rujia Liu 2 #include<cmath> 3 #include<cstdio> 4 #include<iostream> 5 6 using namespace std; 7 8 const double PI = acos(-1.0); 9 const double TWO_PI = 2 * PI; 10 const double eps = 1e-6; 11 12 inline double NormalizeAngle(double rad, double center = PI) { 13 return rad - TWO_PI * floor((rad + PI - center) / TWO_PI); 14 } 15 16 inline int dcmp(double x) { 17 if(fabs(x) < eps) return 0; else return x < 0 ? -1 : 1; 18 } 19 20 struct Point { 21 double x, y; 22 Point(double x=0, double y=0):x(x),y(y) { } 23 }; 24 25 typedef Point Vector; 26 27 inline Vector operator + (Vector A, Vector B) { return Vector(A.x+B.x, A.y+B.y); } 28 inline Vector operator - (Point A, Point B) { return Vector(A.x-B.x, A.y-B.y); } 29 30 inline double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; } 31 inline double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; } 32 inline double Length(Vector A) { return sqrt(Dot(A, A)); } 33 34 // 外接圆圆心。假定三点不共线 35 Point get_circumscribed_center(Point p1, Point p2, Point p3) { 36 double bx = p2.x - p1.x; 37 double by = p2.y - p1.y; 38 double cx = p3.x - p1.x; 39 double cy = p3.y - p1.y; 40 double d = 2 * (bx * cy - by * cx); 41 Point p; 42 p.x = (cy * (bx * bx + by * by) - by * (cx * cx + cy * cy)) / d + p1.x; 43 p.y = (bx * (cx * cx + cy * cy) - cx * (bx * bx + by * by)) / d + p1.y; 44 return p; 45 } 46 47 double DistanceToArc(Point a, Point start, Point mid, Point end) { ///点到圆弧的距离 48 Point p = get_circumscribed_center(start, mid, end); 49 bool CCW = dcmp(Cross(mid - start, end - start)) > 0; 50 double ang_start = atan2(start.y-p.y, start.x-p.x); 51 double ang_end = atan2(end.y-p.y, end.x-p.x); 52 double r = Length(p - start); 53 double ang = atan2(a.y-p.y, a.x-p.x); 54 bool inside; 55 if(CCW) { 56 inside = NormalizeAngle(ang - ang_start) < NormalizeAngle(ang_end - ang_start); 57 } else { 58 inside = NormalizeAngle(ang - ang_end) < NormalizeAngle(ang_start - ang_end); 59 } 60 if(inside) { 61 return fabs(r - Length(p - a)); 62 } 63 return min(Length(a - start), Length(a - end)); 64 } 65 66 int main() { 67 int kase = 0; 68 double x1, y1, x2, y2, x3, y3, xp, yp; 69 while(cin >> x1 >> y1 >> x2 >> y2 >> x3 >> y3 >> xp >> yp) { 70 double ans = DistanceToArc(Point(xp,yp), Point(x1,y1), Point(x2,y2), Point(x3,y3)); 71 printf("Case %d: %.3lf ", ++kase, ans); 72 } 73 return 0; 74 }