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 DescriptionIn a two-dimen=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?
InputThe 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.
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
OutputThe minimum time to travel from A to D, round to two decimals.
Sample Input1 0 0 100 100 0 100 100 2 2 1
Sample Output136.60
Authorlxhgww & momodi
SourceHDOJ Monthly Contest-2010.05.01
Recommendlcy magic three-way method. Is the use of convex.
#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;}