hdu 3400 Line belt
题意：给你两条线段AB，CD；然后给你在AB，CD上的速度P，Q，在其它部分的速度是R，然后求A到D的最短时间。
思路：用三分枚举从AB线段上离开的点，然后再用三分枚举在CD的上的点找到最优点，求距离和时间就可以。
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <cmath> 5 using namespace std; 6 const double eps=1e-8; 7 8 int t; 9 double p,q,r; 10 struct point 11 { 12 double x,y; 13 }a,b,c,d; 14 15 double sqr(double x) 16 { 17 return x*x; 18 } 19 20 double dis(double x1,double y1,double x2,double y2) 21 { 22 return sqrt(sqr(x1-x2)+sqr(y1-y2)+eps); 23 } 24 25 double ok(double t1) 26 { 27 point pos; 28 pos.x=a.x+(b.x-a.x)*((t1*p)/dis(a.x,a.y,b.x,b.y)); 29 pos.y=a.y+(b.y-a.y)*((t1*p)/dis(a.x,a.y,b.x,b.y)); 30 double l1=0,r1=dis(c.x,c.y,d.x,d.y)/q; 31 while(r1-l1>eps) 32 { 33 double mid1=(r1+l1)/2; 34 double mid2=(mid1+r1)/2; 35 double x1=d.x+(c.x-d.x)*((mid1*q)/dis(c.x,c.y,d.x,d.y)); 36 double y1=d.y+(c.y-d.y)*((mid1*q)/dis(c.x,c.y,d.x,d.y)); 37 double x2=d.x+(c.x-d.x)*((mid2*q)/dis(c.x,c.y,d.x,d.y)); 38 double y2=d.y+(c.y-d.y)*((mid2*q)/dis(c.x,c.y,d.x,d.y)); 39 double d1=dis(pos.x,pos.y,x1,y1)/r; 40 double d2=dis(pos.x,pos.y,x2,y2)/r; 41 if(d1+mid1<=d2+mid2) 42 { 43 r1=mid2; 44 } 45 else 46 l1=mid1; 47 } 48 double xx=d.x+(c.x-d.x)*(l1*q)/dis(c.x,c.y,d.x,d.y); 49 double yy=d.y+(c.y-d.y)*(l1*q)/dis(c.x,c.y,d.x,d.y); 50 return dis(pos.x,pos.y,xx,yy)/r+t1+l1; 51 } 52 53 int main() 54 { 55 scanf("%d",&t); 56 while(t--) 57 { 58 scanf("%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y); 59 scanf("%lf%lf%lf%lf",&c.x,&c.y,&d.x,&d.y); 60 scanf("%lf%lf%lf",&p,&q,&r); 61 double ll=0,rr=dis(a.x,a.y,b.x,b.y)/p; 62 while(rr-ll>eps) 63 { 64 double mid1=(ll+rr)/2; 65 double mid2=(mid1+ll)/2; 66 if(ok(mid2)>=ok(mid1)) 67 { 68 ll=mid2; 69 } 70 else 71 rr=mid1; 72 } 73 printf("%.2lf\n",ok(ll)); 74 } 75 }
