Line belt
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1820 Accepted Submission(s): 683
Problem Description In a two-dimensional plane there are two line belts, there are two segments AB and CD, lxhgww’s speed on AB is P and on CD is Q, he can move with the speed R on other area on the plane.
How long must he take to travel from A to D?
Input The first line is the case number T.
For each case, there are three lines.
The first line, four integers, the coordinates of A and B: Ax Ay Bx By.
The second line , four integers, the coordinates of C and D:Cx Cy Dx Dy.
The third line, three integers, P Q R.
0<= Ax,Ay,Bx,By,Cx,Cy,Dx,Dy<=1000
1<=P,Q,R<=10
Output The minimum time to travel from A to D, round to two decimals.
Sample Input 1 0 0 0 100 100 0 100 100 2 2 1
Sample Output 136.60
Author lxhgww&&momodi
Source
HDOJ Monthly Contest – 2010.05.01
Recommend lcy 神奇的三分法。 就是利用凸性。
#include<stdio.h> #include<iostream> #include<math.h> #include<iostream> #include<algorithm> #include<string.h> using namespace std; const double eps=1e-5; struct point { double x,y; }; double dis(point a,point b) { return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); } double p,q,r; double find2(point a,point c,point d) { point left,right; point mid,midmid; double t1,t2; left=c;right=d; do { mid.x=(left.x+right.x)/2; mid.y=(left.y+right.y)/2; midmid.x=(mid.x+right.x)/2; midmid.y=(mid.y+right.y)/2; t1=dis(a,mid)/r+dis(mid,d)/q; t2=dis(a,midmid)/r+dis(midmid,d)/q; if(t1>t2)left=mid; else right=midmid; } while(dis(left,right)>=eps); return t1; } double find(point a,point b,point c,point d) { point left,right; point mid,midmid; double t1,t2; left=a; right=b; do { mid.x=(left.x+right.x)/2; mid.y=(left.y+right.y)/2; midmid.x=(mid.x+right.x)/2; midmid.y=(mid.y+right.y)/2; t1=dis(a,mid)/p+find2(mid,c,d); t2=dis(a,midmid)/p+find2(midmid,c,d); if(t1>t2)left=mid; else right=midmid; }while(dis(right,left)>=eps); return t1; } int main() { int T; point a,b,c,d; scanf("%d",&T); while(T--) { scanf("%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y); scanf("%lf%lf%lf%lf",&c.x,&c.y,&d.x,&d.y); scanf("%lf%lf%lf",&p,&q,&r); printf("%.2lf\n",find(a,b,c,d)); } return 0; }