ACM ICPC 2014 NEERC (Northeastern European Regional Contest) Problems Review

1. ACM ICPC 2014{2015 Northeastern European Regional Contest Problems Review Roman Elizarov December 7, 2014

2. Problem A. Alter Board I The minimal answer to this problem is bn=2c + bm=2c I The solution is to make inversions on each even row and each even column I To prove that the answer is minimal consider the

3. rst column with its n cells that form n 1 neighbouring pairs I to turn all cells of the

4. rst column in the same color inversions must span the

5. rst column I each spanning inversion makes at most two neighbouring pairs of the same color I so the minimum of d(n 1)=2e = bn=2c inversions are needed I Then consider the top row in the same way

6. Problem B. Burrito King I Consider the problem as a sum of vectors in (a; b) coordinates I The resulting vector may not go above b = B line and must extend on a axis as far as possible I It is optimal to greedily add (ai ; bi ) ingredient vectors starting from the ones that have the least angle to 0a line (or maximal ai=bi ), until b = B line is crossed I Be careful about corner cases with ai = 0 and/or bi = 0 a b B 0 A

7. Problem C. Cactus Generator I This is a straightforward problem for parsing and OO design I De

8. ne class for graph with a method to generate graph given index of the

9. rst and the vertices I De

10. ne class for various range types I Parse and construct classes tree I Build the resulting graph I Connect arbitrary pairs of vertices of odd degree in the resulting graph using temporary edges I Use classical algorithm for Eulerian path I Remove temporary edges to get the minimal number of covering paths

11. Problem D. Damage Assessment I Numerically integrate the square section by dx I The square of the cut at a given x coordinate is a simple planar geometry problem I Take care about leftmost point with in

12. nite derivative I however, the required precision does not make this a big problem I the square section at this point is small x

13. Problem E. Epic Win! I There is a simple solution with up to n2 states I Build your FSM as n copies of a winning FSM with n states I Each state of a winning FSM corresponds to a state in the opponent FSM I Each move of a winning FSM is a winning move for the corresponding opponent's move I Next state in a winning FSM corresponds to the opponent move and opponent's next state I Leave other transitions unde

14. ned I The

15. rst copy of a winning FMS starts in its

16. rst state and wins an opponent that stats in it

17. rst state by construction I Model the behaviour of the opponent and your FSM for all opponent start states from the states 2 to n I When a yet unde

18. ned transition is reached, then insert a transition to a fresh copy of a winning FSM into the state corresponding to the opponent's, thus ensuring win in this copy I Stop modelling when loop is detected I Loop is inside one copy of a winning FSM and is always winning by construction

19. Problem F. Filter I Nothing fancy here I Just implement what the problem statement asks for in a straightforward way I The hardest part seems to be reading and understanding the problem statement

20. Problem G. Gomoku I The

21. rst player's strategy has pretty strict priorities in the moves it makes and it can be exploited I Make the

22. rst move into the free space of the board 2 1 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 15 16 16 17 17 18 18 19 19

23. Problem G. Gomoku cont'd I The opponent must play around the center and you form a diagonal 2 4 3 1 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 15 16 16 17 17 18 18 19 19

24. Problem G. Gomoku cont'd I The opponent forms three in a row and you make defensive moves 2 4 6 3 1 5 7 8 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 15 16 16 17 17 18 18 19 19

25. Problem G. Gomoku cont'd I The opponent closes two in a row at one side, and you extend in on the other 10 2 4 9 6 3 1 5 7 8 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 15 16 16 17 17 18 18 19 19

26. Problem G. Gomoku cont'd I The opponent closes the three on the other side, but you continue oence at building a winning position 11 10 2 4 12 9 6 3 1 5 7 8 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 15 16 16 17 17 18 18 19 19

27. Problem G. Gomoku cont'd I Force the opponent into a sequence of defensive moves I Then close four in a row with a hole that is formed by the opponent defence 15 11 14 18 10 19 2 20 4 16 12 17 9 13 6 3 1 5 7 8 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 15 16 16 17 17 18 18 19 19

28. Problem G. Gomoku cont'd I The opponent closes your open three, you extend it, forming a winning fork 15 22 11 14 18 10 19 2 20 4 16 12 17 9 21 13 6 3 1 5 7 8 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 15 16 16 17 17 18 18 19 19

29. Problem G. Gomoku win I You win I It is very hard to win otherwise, because playing

30. rst in gomoku gives an enormous advantage even to such a simple strategy 26 23 15 22 11 14 18 10 19 2 20 4 16 12 17 24 9 21 13 25 6 3 1 5 7 8 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 15 16 16 17 17 18 18 19 19

31. Problem H. Hidden Maze I Make a rooted tree I Lets compute how many times each edge is a median I Start with an edge with lowest ci and work in increasing order of ci I For each edge ci look at its lowest vertex j in the tree I For each path from j down into the subtree, let the balance be the number of edges with c higher than current ci minus the number of edges with c lower than current ci I For each vertex j maintain an array bj I with 2dj + 1 elements bj [] for jj di , where dj is a depth of subtree rooted at j I each item bj [] contains a number of paths down from j with a balance I including an empty path with balance zero

32. Problem H. Hidden Maze cont'd I Initial bj [] is the number of paths of a length down from vertex j P I It is easy to compute recursively in O( dj ) while building rooted tree I From the current vertex j walk up the tree I For all vertices k up tree from j compute the number of paths with balance zero going from down up to j , then up to k then down to other subtree of k I paths with zero balance are the ones where ci is the median X =dj :::dj bj [] (bk [ k;j] bk#[ k;j k;k#]) I where k # is the next vertex from k down on the path to j and j is the next vertex up from j I and k;j is the sum of balances on a path from k to j I The total complexity is O( P dj hj ), where hj is the height of vertex j | length of path from root

33. Problem H. Hidden Maze cont'd I Update bj [] when done with an edge ci I For all vertices k up tree from j update the bk arrow taking into account that ci balance changes from 1 to 1 bk [] bk [] + bj [ k;j + 1] bj [ k;j 1] I The total complexity is also O( P dj hj ) I However, for the graph randomly generated as described in the problem statement P (dj hj ) = O(n p n)

34. Problem I. Improvements I Consider transposition aj | the number of ship at coordinate j , that is reverse to what is given in the input I It is easy to prove that the chain of ships that remain on their initial position corresponds to a subsequence of aj with a special property: I it is an increasing sequence of numbers aj followed by decreasing sequence of numbers aj I Increasing/decreasing subsequence is a well-known problem with O(n log n) solution using dynamic programming 0 0 1 1 3 2 4 3 2 4

35. Problem J. Jokewithpermutation I This problem is solved with exhaustive search I for each number try all positions that it can occupy I start search with numbers that can occupy fewest number of possible positions

36. Problem K. Knockout Racing I Nothing fancy here I Just implement what the problem statement asks for in a straightforward way I This is the easiest problem in the contest

ACM ICPC 2014 NEERC (Northeastern European Regional Contest) Problems Review

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ACM ICPC 2014 NEERC (Northeastern European Regional Contest) Problems Review