OJ Board

Since you have found the equation and you can solve it using little mathematics. You need only one line to produce the correct answer, don't you?

yatsen wrote:Can you post your solution?
Thanks!

He is referring to another problem "Power Transmission."
Not this one.

Thanks Yarin! I've solved it today using this different approach:

Greed Algorithm: at each step choose the point of interest not covered which is adjacent to the biggest amount of points of interests which are not yet covered. If there is more than one point of interest with the same minimum ...

How do you send your program to the judge?
This may be a problem.

Output "0 rings" instead of "0 ring". This is a matter of English grammar.

Hi,
Can someone explain the solution to this problem ?
it looks similar to the NIM-game.

Thank you.
Convert all numbers into binary sequences and list them all one in a row.
Then you may think what the best strategy is for playing the game.
If this is still obscure, consider the special case when ...

The airplane must land on arrival, the altitude of it thus must be zero at the destination.

Finally I got AC!
Thanks for your great help!

Adrian Kuegel wrote:Yes, I think you misunderstood the description. The 50% is 50% of the number of roads, not of the length.
Try this input:
29 10 10
Output:
225 2303

The problem says "there can be at most 50% circular roads and the
number of them will be no more than straight ones."
So let the number of circular roads be x and the length of a circular road
be cl, I have the following formula:

x*cl/(x*cl+(e-x)*dist) <= 0.5

after some algebra, we can get x ...

I fixed my codes, but still wrong answer......

What's wrong with my codes???
Is my formula wrong???
Or another precision error???

If you consider the problem ,which is about Root instead of Bishop,
maybe it is easy.
now ,you could rotate the board by 45 degrees , then youn could consider the Bishop like a Root.
Good lucK!!
hmm...I rotate the board by 45 degrees and then don't know what to do next...
Can you explain how to ...

Your algorithm is wrong, check your inner loop.
And there is no need to use big numbers.

Yeah! I think you are right.
This problem can be formulated as a 2-SAT problem
and use dynamic programming to solve it. Thanx

I try to formulate this problem as a two-coloring problem.
My idea is that I treat each pair of input point as a single node.
And for each pair of node, if their desired paths may conflict, they are
joined by an edge. Then I just check this graph to see if it is two-colorable.
But this idea seems ...