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вернуться в форумI always get WA 2...., and I don't know why...... My program is full search(rather primitive) but got Ac.
Next tests also simple:
1 1
2 2
1 -3
2 -6
0.00000000
5 5
10 5
-1 -4
-3 -4
1.581139
-100 1
-200 1
-100 2
-200 2
2.0000000
-1 0
1 -2
-2 0
3 -10
0.485071
Edited by author 27.10.2009 15:49 Could you tell me how you get 2-nd answer(1.581139)
Is it a distance from starting point to the lines intersrction?
If it is, could you give me the coordinates of lines intersection?
P.S. I got 1.597191412
I depend on obviouse statement:
optimal path is some segment from (0,0) to
one of lines and projection to another one.
So I make full search about this situation.
You think about line inersection as one of
solutions of variational equation.
But there are many other solutions. In my solution I also try this kind of optimal path, but I couldn't get your answer 1.581139 ..... It is simple. Soon I will give to this
issue two points: A2 abd A3( A1=(0,0)) and this will be
best help. 5 5
10 5
-1 -4
-3 -4
1.581139
Intersection (0.7142857,-1.4285714).
First move to (0.84995501, -1.22506748)
(Too far going to intersection, we are crossing angle.)
Edited by author 28.10.2009 09:10 How did you calculate the point of first move? I got 1.5811388 too, but first point is (0.849057,-1.22642), which lies on a line p:=((5,5),(-1,-4)). I searched such points with a help of finding a conditional extremal value of a function: F(x,y):=Dist((0,0),point)+Dist(point,q) (q is the second line), where point lies on p, and a function G(x,y):=Dist((0,0),point)+Dist(point,p), where point lies on q.
But I still have WA#3((( I don't know any reason.
Edited by author 31.10.2009 03:13
Edited by author 31.10.2009 04:07 |
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