csu 1503: 点到圆弧的距离

1503: 点到圆弧的距离

Time Limit: 1 Sec  Memory Limit: 128 MB  Special Judge
Submit: 614  Solved: 101
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Description

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。

Input

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。

Output

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。你的输出和标准输出之间最多能有0.001的误差。

Sample Input

0 0 1 1 2 0 1 -1
3 4 0 5 -3 4 0 1

Sample Output

Case 1: 1.414
Case 2: 4.000

HINT

Source

湖南省第十届大学生计算机程序设计竞赛

CSU后台出了问题,找了标准输入输出文件,找不出差异。测试数据下载

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这个题磨死我了....学长们比赛的时候是怎么AC的...

说下我的思路:我的思路是判断 弧_ABC 为劣弧还是优弧,怎么判断 弧_ABC是劣弧还是优弧呢?我们可以判断扇形 S_AOC 是否等于 S_BOC+S_AOB 如果相等,那么 ABC 就是劣弧,否则就是优弧.这里要特判一下平角,不然结果会有问题。如果 弧_ABC 为劣弧,我们只要求出射线OP和圆的交点P1,那么只要 S_AOP1+S_P1OC = S_AOC，那么P就在弧ABC里面,最短距离就是|OP-r|.否则,P就没有在弧ABC内,最短距离就是min(AP,CP).如果弧_ABC为优弧,同样求出S_AOC，S_AOP1+S_P1OC如果 S_AOP1+S_P1OC = S_AOC，那么P就不在弧ABC里面,最短距离就是min(AP,CP).否则,P就在弧ABC内,最短距离就是|OP-r|.如果是角AOC是平角,那么我们就要特判.判断OP和OB是否是否在OC的同一侧(这里可以用叉积判断),在同一侧，距离就是|OP-r|,否则就是min(AP,CP)...还有这个题最坑的一点,幸亏有测试用例,当我们求圆心角时,我是用 a = acos(OA*OB / |OA*OB|) ，但是由于精度原因,acos()有可能不属于 [-1,1]，但是即使误差小,但是计算机也是会算错的,所以就算是acos(1.00000000000000000200)这样求出的结果也是NAN。所以要强制的转换成1，-1.这样的话整个程序就完成了.

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#define eps (1e-4)
#define N 205
#define dd double
#define sqr(x) ((x)*(x))
const double pi = acos(-);
using namespace std;
struct Point
{
    double x,y;
};
double cross(Point a,Point b,Point c) ///叉积
{
    return (a.x-c.x)*(b.y-c.y)-(b.x-c.x)*(a.y-c.y);
}
double dis(Point a,Point b) ///距离
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double mult(Point a,Point b,Point c)  ///点积
{
    return (a.x-c.x)*(b.x-c.x)+(a.y-c.y)*(b.y-c.y);
}
Point waixin(Point a,Point b,Point c) ///外接圆圆心坐标
{
    Point p;
    double a1 = b.x - a.x, b1 = b.y - a.y, c1 = (a1*a1 + b1*b1)/;
    double a2 = c.x - a.x, b2 = c.y - a.y, c2 = (a2*a2 + b2*b2)/;
    double d = a1*b2 - a2*b1;
    p.x = a.x + (c1*b2 - c2*b1)/d, p.y=a.y + (a1*c2 -a2*c1)/d;
    return p;
}
Point intersection(Point a,Point b,Point c,Point d)
{
    Point p = a;
    double t =
        ((a.x-c.x)*(c.y-d.y)-(a.y-c.y)*(c.x-d.x))/((a.x-b.x)*(c.y-d.y)-(a.y-b.y)*(c.x-d
                .x));
    p.x +=(b.x-a.x)*t;
    p.y +=(b.y-a.y)*t;
    return p;
}
void intersection_line_circle(Point c,double r,Point l1,Point l2,Point& p1,Point& p2)
{
    Point p=c;
    double t;
    p.x+=l1.y-l2.y;
    p.y+=l2.x-l1.x;
    p=intersection(p,c,l1,l2);
    t=sqrt(r*r-dis(p,c)*dis(p,c))/dis(l1,l2);
    p1.x=p.x+(l2.x-l1.x)*t;
    p1.y=p.y+(l2.y-l1.y)*t;
    p2.x=p.x-(l2.x-l1.x)*t;
    p2.y=p.y-(l2.y-l1.y)*t;
}
double getarea(Point a,Point b,Point c,double r){ ///求扇形面积
    double temp = mult(a,b,c)/(dis(a,c)*dis(b,c));
    if(temp<=-) temp = -; ///这里一定要记得判范围..不然 1.00000000000000000200 得到的acos()也是 NAN
    if(temp>=) temp = ;
    double angle = acos(temp);
    return fabs(angle*r*r/);
}
int main()
{
    Point a,b,c,p,p1,p2,p3;
    int t = ;
    //freopen("a.in","r",stdin);
    //freopen("a.txt","w",stdout);
    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)
    {
        Point circle = waixin(a,b,c);
        double r = dis(circle,a);
        double ans = min(dis(p,a),dis(p,c));
        double op = dis(circle,p);
        double area1 = getarea(a,c,circle,r);
        double area2 = getarea(a,b,circle,r);
        double area3 = getarea(b,c,circle,r);
        intersection_line_circle(circle,r,circle,p,p1,p2);
        if(acos(mult(p,p1,circle)/((dis(p,circle))*dis(p1,circle)))<) p3 = p2;
        else p3 = p1;
        double area4 = getarea(a,p3,circle,r);
        double area5 = getarea(c,p3,circle,r);
        double temp = mult(a,c,circle)/(dis(a,circle)*dis(c,circle));
        if(temp<=-) temp = -;
        if(temp>=) temp = ;
        double angle_aoc = acos(temp);
        if(fabs(angle_aoc-pi)<eps){ ///平角特殊处理
            if(cross(p,c,circle)*cross(b,c,circle)>=){
                ans = min(ans,fabs(op-r));
            }
            printf("Case %d: %.3lf\n",t++,ans);
            continue;
        }
        if(fabs(area1-area2-area3)<eps)  ///劣弧
        {
            if(fabs(area1-area4-area5)<eps) ans = min(ans,fabs(op-r));
        }
        else
        {
            if(fabs(area1-area4-area5)>eps) ans = min(ans,fabs(op-r));
        }
        printf("Case %d: %.3lf\n",t++,ans);
    }
}

解法二:学校队友提供(利用弧度求解)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#define PI acos(-1.0)
using namespace std;
double r,ans,dis;
double ant,ant1,ant2,ant3;
struct Point
{
    double x,y;
}p[];
double jisuan(Point a,Point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
bool is_ni()
{
    if(ant3>ant1)
    {
        if(ant2<ant3&&ant2>ant1)return false;
        else return true;
    }
    if(ant3<ant1)
    {
        if(ant2<ant1&&ant2>ant3)return true;
        return false;
    }
}
void get_fang()
{
     if(p[].y>)
     {
         if(p[].x==)ant=PI/;
         else ant=atan2(p[].y,p[].x);
     }
     else if(p[].y==)
     {
         if(p[].x>)ant=;
         else ant=PI;
     }
     else if(p[].y<)
     {
         if(p[].x==)ant=*PI/;
         ant=*PI+atan2(p[].y,p[].x);
     }
     if(p[].y>)
     {
         if(p[].x==)ant1=PI/;
         else ant1=atan2(p[].y,p[].x);
     }
     else if(p[].y==)
     {
         if(p[].x>)ant1=;
         else ant1=PI;
     }
     else if(p[].y<)
     {
         if(p[].x==)ant1=*PI/;
         ant1=*PI+atan2(p[].y,p[].x);
     }
     if(p[].y>)
     {
         if(p[].x==)ant2=PI/;
         else ant2=atan2(p[].y,p[].x);
     }
     else if(p[].y==)
     {
         if(p[].x>)ant2=;
         else ant2=PI;
     }
     else if(p[].y<)
     {
         if(p[].x==)ant2=*PI/;
         ant2=*PI+atan2(p[].y,p[].x);
     }
     if(p[].y>)
     {
         if(p[].x==)ant3=PI/;
         else ant3=atan2(p[].y,p[].x);
     }
     else if(p[].y==)
     {
         if(p[].x>)ant3=;
         else ant3=PI;
     }
     else if(p[].y<)
     {
         if(p[].x==)ant3=*PI/;
         ant3=*PI+atan2(p[].y,p[].x);
     }
}
Point getyuanxin(Point p1, Point p2, Point p3)
{
    double a, b, c, d, e, f;
    Point p;
    a = *(p2.x-p1.x);
    b = *(p2.y-p1.y);
    c = p2.x*p2.x+p2.y*p2.y-p1.x*p1.x-p1.y*p1.y;
    d = *(p3.x-p2.x);
    e = *(p3.y-p2.y);
    f = p3.x*p3.x+p3.y*p3.y-p2.x*p2.x-p2.y*p2.y;
    p.x = (b*f-e*c)/(b*d-e*a);
    p.y = (d*c-a*f)/(b*d-e*a);
    r = sqrt((p.x-p1.x)*(p.x-p1.x)+(p.y-p1.y)*(p.y-p1.y));
    return p;
}
int main()
{  // freopen("a.in","r",stdin);
  //  freopen("out.txt","w",stdout);
    int l=;
    while(~scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&p[].x,&p[].y,&p[].x,&p[].y,&p[].x,&p[].y,&p[].x,&p[].y))
    {   int flag=,flag1=;
        p[]=getyuanxin(p[],p[],p[]);
        dis=jisuan(p[],p[]);
        r=jisuan(p[],p[]);
        for(int i=;i<=;i++)p[i].x-=p[].x,p[i].y-=p[].y;
        get_fang();
        if(is_ni())
        {
            if(ant1>ant2)
            {
                if(ant>=ant1||ant<=ant2)ans=abs(dis-r);
                else ans=min(jisuan(p[],p[]),jisuan(p[],p[]));
            }
            else
            {
                if(ant<=ant2&&ant>=ant1)ans=abs(dis-r);
                else ans=min(jisuan(p[],p[]),jisuan(p[],p[]));
            }
        }
        else
        {
            if(ant1>ant2)
            {
                if(ant>=ant2&&ant<=ant1)ans=abs(dis-r);
                else ans=min(jisuan(p[],p[]),jisuan(p[],p[]));
            }
           else
            {
                if(ant<=ant1||ant>=ant2)ans=abs(dis-r);
                else ans=min(jisuan(p[],p[]),jisuan(p[],p[]));
            }
        }
        printf("Case %d: %.3lf\n",l++,ans);
    }
    return ;
}

解法三:经过冥思苦想,终于想到了怎样判断一个点是否在两个向量的夹角之间!

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L6D+iFHkx/fAdAAAAAAElFTkSuQmCC" alt="" width="246" height="208" />

大家请看上图,假设当前 OA 在 OC 的逆时针方向,现在B点绕C点旋转相对于O点是顺时针方向,B点绕A点的话相对于O点是逆时针方向,然后这样就可以判断B点是否在OA~OB之间了，然后马上就可以判断当前弧是优弧还是劣弧,这样的话P点也以同样的方式判断就可以得到其位于优弧还是劣弧了(只不过P点要判0)，但是还是要特判平角.不过平角的判断的判断我也做了一个小小的优化,不用角度判了,只用 2*r == dis(a,c) 判断就OK，这样就不怕被角度坑了.

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#define eps (1e-4)
#define N 205
#define dd double
#define sqr(x) ((x)*(x))
const double pi = acos(-);
using namespace std;
struct Point
{
    double x,y;
};
double cross(Point a,Point b,Point c) ///叉积
{
    return (a.x-c.x)*(b.y-c.y)-(b.x-c.x)*(a.y-c.y);
}
double dis(Point a,Point b) ///距离
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
Point waixin(Point a,Point b,Point c) ///外接圆圆心坐标
{
    Point p;
    double a1 = b.x - a.x, b1 = b.y - a.y, c1 = (a1*a1 + b1*b1)/;
    double a2 = c.x - a.x, b2 = c.y - a.y, c2 = (a2*a2 + b2*b2)/;
    double d = a1*b2 - a2*b1;
    p.x = a.x + (c1*b2 - c2*b1)/d, p.y=a.y + (a1*c2 -a2*c1)/d;
    return p;
}
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)
    if(cross(b,c,a)>&&cross(p,a,c)<&&cross(p,a,b)>){ ///b 在 c 的逆时针方向
        return true;
    }
    if(cross(b,c,a)<&&cross(p,a,c)>&&cross(p,a,b)<){ ///b 在 c 的顺时针方向
        return true;
    }
    return false;
}
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)
    if(cross(b,c,a)>&&cross(p,a,c)<=&&cross(p,a,b)>=){ ///b 在 c 的逆时针方向
        return true;
    }
    if(cross(b,c,a)<&&cross(p,a,c)>=&&cross(p,a,b)<=){ ///b 在 c 的顺时针方向
        return true;
    }
    return false;
}
int main()
{
    Point a,b,c,p;
    int t = ;
    freopen("a.in","r",stdin);
    freopen("a.txt","w",stdout);
    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)
    {

        Point circle = waixin(a,b,c);
        double r = dis(circle,a);
        double ans = min(dis(p,a),dis(p,c));
        double op = dis(circle,p);
        if(fabs(*r-dis(a,c))<eps){ ///平角特殊处理
            if(cross(p,c,circle)*cross(b,c,circle)>=){
                ans = min(ans,fabs(op-r));
            }
            printf("Case %d: %.3lf\n",t++,ans);
            continue;
        }
        if(judge(circle,a,c,b))  ///劣弧
        {
            if(judge1(circle,a,c,p)) ans = min(ans,fabs(op-r));
        }
        else
        {
            if(!judge(circle,a,c,p)) ans = min(ans,fabs(op-r));
        }
        printf("Case %d: %.3lf\n",t++,ans);
    }
}

官方标程:

// Rujia Liu
#include<cmath>
#include<cstdio>
#include<iostream>

using namespace std;

const double PI = acos(-1.0);
const double TWO_PI =  * PI;
const double eps = 1e-;

inline double NormalizeAngle(double rad, double center = PI) {
  return rad - TWO_PI * floor((rad + PI - center) / TWO_PI);
}

inline int dcmp(double x) {
  if(fabs(x) < eps) return ; else return x <  ? - : ;
}

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

typedef Point Vector;

inline Vector operator + (Vector A, Vector B) { return Vector(A.x+B.x, A.y+B.y); }
inline Vector operator - (Point A, Point B) { return Vector(A.x-B.x, A.y-B.y); }

inline double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }
inline double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }
inline double Length(Vector A) { return sqrt(Dot(A, A)); }

// 外接圆圆心。假定三点不共线
Point get_circumscribed_center(Point p1, Point p2, Point p3) {
  double bx = p2.x - p1.x;
  double by = p2.y - p1.y;
  double cx = p3.x - p1.x;
  double cy = p3.y - p1.y;
  double d =  * (bx * cy - by * cx);
  Point p;
  p.x = (cy * (bx * bx + by * by) - by * (cx * cx + cy * cy)) / d + p1.x;
  p.y = (bx * (cx * cx + cy * cy) - cx * (bx * bx + by * by)) / d + p1.y;
  return p;
}

double DistanceToArc(Point a, Point start, Point mid, Point end) {   ///点到圆弧的距离
  Point p = get_circumscribed_center(start, mid, end);
  bool CCW = dcmp(Cross(mid - start, end - start)) > ;
  double ang_start = atan2(start.y-p.y, start.x-p.x);
  double ang_end = atan2(end.y-p.y, end.x-p.x);
  double r = Length(p - start);
  double ang = atan2(a.y-p.y, a.x-p.x);
  bool inside;
  if(CCW) {
    inside = NormalizeAngle(ang - ang_start) < NormalizeAngle(ang_end - ang_start);
  } else {
    inside = NormalizeAngle(ang - ang_end) < NormalizeAngle(ang_start - ang_end);
  }
  if(inside) {
    return fabs(r - Length(p - a));
  }
  return min(Length(a - start), Length(a - end));
}

int main() {
  int kase = ;
  double x1, y1, x2, y2, x3, y3, xp, yp;
  while(cin >> x1 >> y1 >> x2 >> y2 >> x3 >> y3 >> xp >> yp) {
    double ans = DistanceToArc(Point(xp,yp), Point(x1,y1), Point(x2,y2), Point(x3,y3));
    printf("Case %d: %.3lf\n", ++kase, ans);
  }
  return ;
}