HDU Line Belt

The problem is a 3-point search. In fact, this problem often appears in high school analytic geometry, the idea is very simple, you can see that the AB line and the CD segment on the existence of a point x and Y to make the time spent least

Because the time on the AB and the CD has a convex function with the coordinates of the X and Y points, you can think of the solution using a 3-point search method. Of course, we need to use two three-point search nesting, lock x after finding the y that satisfies the condition, and know the smallest x and y that satisfy the condition. The advantage of using a three-point search is

The midpoint coordinate formula of two points can be used, which is more convenient than solving with the mathematical formula.

#include"iostream"#include"stdio.h"#include"algorithm"#include"string.h"#include"Cmath"#defineMX 1005#defineExp 1e-6using namespacestd;DoubleP,q,r;structpoint{Doublex, y;};DoubleDist (Point A,point b) {returnsqrt ((a.x-b.x) * (a.x-b.x) + (A.Y-B.Y) * (a.y-b.y));}DoubleFind2 (Point x,point c,point d) {point left,right,mid,midmid; Doublet1,t2; Left=B; 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=dist (X,mid)/r+dist (mid,d)/Q; T2=dist (X,midmid)/r+dist (midmid,d)/P; if(T1>T2) left=mid; Elseright=Midmid; } while(ABS (T1-T2) >exp); returnT1;}DoubleFind1 (Point a,point b,point c,point d) {point left,right,mid,midmid; Doublet1,t2; Left=A; Right=b;  Do //Loop First, so you don't have to assign an initial value to T1,t2{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=dist (A,mid)/p+Find2 (mid,c,d); T2=dist (A,midmid)/p+Find2 (midmid,c,d); if(T1>T2) left=mid; Elseright=Midmid; } while(ABS (T1-T2) >exp); returnT1;}intMain () {intT; CIN>>T;    Point A,b,c,d;  while(t--) {cin>>a.x>>a.y>>b.x>>b.y>>c.x>>c.y>>d.x>>D.y; CIN>>p>>q>>R; printf ("%.2lf\n", Find1 (a,b,c,d)); }    return 0;}

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HDU Line Belt