标签:style class blog c code http
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今天无事水一水,结果就看到这个水题了!
题意思是 有俩个区域如图

求在俩个圆之间的运动时间 给出 初始的开始点和速度的矢量式;而且这个点 不再俩个圆之间的区域,且碰到内测员会反弹:
其实就是求 与俩个圆的交点! 代码如下:()
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#include <cstring>#include <cmath>#include <algorithm>#include <cstdlib>#include <cstdio>using
namespace std;struct
point{ double
x,y; point (double
x=0,double
y=0):x(x),y(y){} void
show() { printf("%lf %lf\n",x,y); }};typedef
point Vector;const
double eps=1e-8;int
dcmp(double
x){ if(fabs(x)<eps) return
0; return
x<0?-1:1;}Vector operator + (Vector a,Vector b){return
Vector(a.x+b.x,a.y+b.y);}Vector operator - (Vector a,Vector b){return
Vector(a.x-b.x,a.y-b.y);}Vector operator * (Vector a,double
b){return
Vector(a.x*b,a.y*b);}double
operator / (Vector a,Vector b){if(dcmp(b.x)!=0)return
a.x/b.x;return
a.y/b.y;}double
det(Vector a,Vector b){return
a.x*b.y-a.y*b.x;}double
dot(Vector a,Vector b){return
a.x*b.x+a.y*b.y;}double
lenth(Vector a){return
sqrt(dot(a,a));}struct
line{ point p; Vector v; double
angle; line(){} line(point p,Vector v):p(p),v(v){} bool
operator <(const
line &rht)const { return
angle<rht.angle; }};struct
circle{ point c; double
r; circle(){c=point(0.0,0.0);} circle(point c,double
r):c(c),r(r){} point Point(double
rad) { return
point(c.x+cos(rad)*r,c.y+sin(rad)*r); }};int
get_circle_intersection(line L,circle C,double
&t1,double
&t2){ t1=t2=0; double
a=L.v.x, b=L.p.x-C.c.x,c=L.v.y,d=L.p.y-C.c.y; double
e=a*a+c*c,f=2*(a*b+c*d),g=b*b+d*d-C.r*C.r; double
detle = f*f-4*e*g; if(dcmp(detle)<0) return
0; if(dcmp(detle)==0) {t1=t2=-f/(2*e);return
1;} t1=(-f-sqrt(detle)) /(2*e); t2=(-f+sqrt(detle)) /(2*e); if(dcmp(t1)<0 || dcmp(t2)<0) return
0;//按照速度的反方向才可以 return
2;}int
main(){ double
t1,t2; double
x1,x2; line tmp; circle tmp1; circle tmp2; double
Rm, R, r; while(~scanf("%lf %lf %lf %lf %lf %lf %lf",&Rm,&R,&r,&tmp.p.x,&tmp.p.y,&tmp.v.x,&tmp.v.y)) { Rm+=r;R+=r; tmp1.r=Rm; tmp2.r=R; int
count1=get_circle_intersection(tmp,tmp1,t1,t2); int
count2=get_circle_intersection(tmp,tmp2,x1,x2); if(count2==0)printf("0.00\n"); else printf("%.3lf\n",fabs(x2-x1)-fabs(t1-t2));// 因为直线式点+向量(和速度一样)所以减法就可以了 } return
0;} |
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标签:style class blog c code http
原文地址:http://www.cnblogs.com/shuly/p/3751058.html