题意: 给一个点,一个圆,一个矩形, 求一条折线,从点出发,到圆,再到矩形的最短距离。
解法: 因为答案要求输出两位小数即可,精确度要求不是很高,于是可以试着爆一发,暴力角度得到圆上的点,然后求个距离,求点到矩形的距离就是求点到四边的距离然后求个最小值,然后总的取个最小值即可。
代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#define Mod 1000000007
#define pi acos(-1.0)
#define eps 1e-8
using namespace std;
#define N 1000107
typedef struct Point
{
double x,y;
Point(double x=0,double y=0):x(x),y(y){}
Point(){}
}Vector;
int sgn(double x)
{
if(x > eps) return 1;
if(x < -eps) return -1;
return 0;
}
Vector operator + (Vector A,Vector B) { return Vector(A.x+B.x,A.y+B.y); }
Vector operator - (Point A,Point B) { return Vector(A.x-B.x,A.y-B.y); }
bool operator == (const Point& a,const Point& b) { return (sgn(a.x-b.x) == 0 && sgn(a.y-b.y) == 0); }
double Dot(Vector A,Vector B) { return A.x*B.x + A.y*B.y; }
double Cross(Vector A,Vector B) { return A.x*B.y - A.y*B.x; }
double Length(Vector A) { return sqrt(Dot(A,A)); }
double PtoSeg(Point P,Point A,Point B)
{
if(A == B) return Length(P-A);
Vector V1 = B-A;
Vector V2 = P-A;
Vector V3 = P-B;
if(sgn(Dot(V1,V2)) < 0) return Length(V2);
else if(sgn(Dot(V1,V3)) > 0) return Length(V3);
else return fabs(Cross(V1,V2))/Length(V1);
}
int main()
{
double X,Y;
double Cx,Cy,R;
double x1,y1,x2,y2;
while(scanf("%lf%lf",&X,&Y)!=EOF)
{
if(sgn(X) == 0 && sgn(Y) == 0) break;
scanf("%lf%lf%lf",&Cx,&Cy,&R);
scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
Point A = Point(x1,y1);
Point B = Point(x1,y2);
Point C = Point(x2,y1);
Point D = Point(x2,y2);
double delta = 2.0*pi*0.0001;
double Min = Mod;
for(int Angle=1;Angle<=10000;Angle++)
{
double ang = delta*Angle;
double nx = Cx + R*cos(ang);
double ny = Cy + R*sin(ang);
double D1 = PtoSeg(Point(nx,ny),A,B);
double D2 = PtoSeg(Point(nx,ny),A,C);
double D3 = PtoSeg(Point(nx,ny),B,D);
double D4 = PtoSeg(Point(nx,ny),C,D);
double Dis = sqrt((nx-X)*(nx-X)+(ny-Y)*(ny-Y))+min(min(min(D1,D2),D3),D4);
Min = min(Dis,Min);
}
printf("%.2f
",Min);
}
return 0;
}