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Houyang

Houyang. 3/25/13. USACO March Contest. Congrats to Johnny Ho for scoring over 900 points in the Gold Division 7th place in US Kudos to Jonathan Uesato too, for a score of 878 Almost slayed the Ho. Bronze: Breed Proximity. N (1 <= N <= 50000) cows in a line.

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Houyang

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  1. Houyang 3/25/13

  2. USACO March Contest • Congrats to Johnny Ho for scoring over 900 points in the Gold Division • 7th place in US • Kudos to Jonathan Uesato too, for a score of 878 • Almost slayed the Ho

  3. Bronze: Breed Proximity • N (1 <= N <= 50000) cows in a line. • Each has a breed (breeds are integers 0...1000000) • Cows can be crowded: • Two cows of the same breed are said to be "crowded" if their positions within the line differ by no more than K (1 <= K <= N) • Find the maximum breed of a pair of crowded cows. • Sample: N = 6, K = 3 • Breeds: 7 3 4 2 3 4 • Output: 4

  4. Bronze: Breed Proximity Solution • Iterate over the cows from left to right. • For each breed, store the position of the last cow of that breed. • Use an array, because 1000000 is small enough. • Could also use a map. Runtime: O(N)

  5. Silver/Gold: The Cow Run • N (1 <= N <= 1000) cows on a number line. • Each has a coordinate P (|P| <= 500000) • Each minute, a rampaging cow causes one dollar worth of damage. • Farmer John can walk over to a cow and calm it. • He starts at the origin and walks 1 unit per minute. • Farmer John wants to minimize the total damage. Compute the minimum. • Sample: N = 4 • Positions: -2, -12, 3, 7 • Output: 50 (visit -2, then 3, then 7, then -12)

  6. Silver/Gold: The Cow Run Solution • Separate the cows into two arrays: positive / negative coordinates. • Sort each. Then "merge" the two arrays, visiting the next positive or next negative cow each step. After N steps, all cows will be calmed. • DP state: #negative cows, #positive cows, and a flag to represent FJ's location • If flag is 1, FJ is on the positive side. • Else, FJ is on the negative side. • This is O(N^2).

  7. Gold: Necklace • Bessie has string A of length N (N <= 10000) and string B of length M (M <= 1000) • Bessie wants to remove some characters from A so that B is not a substring of A. • Find the minimum number of characters to remove. • Sample: A = ababaa, B = aba • Output: 1. (remove an a to make abbaa)

  8. Gold: Necklace Solution • DP State: Position in A, Position in B • To transition in amortized O(1), use a KMP string matching algorithm failure function table. • Each step, you can either keep the character in A, or delete it. • If you delete it, Position in B doesn't change. • If you keep it, you compute the new position in B using the table. • O(NM)states.

  9. POTW • You have N nodes, M bidirectional edges. • How many groups of 3 nodes are all connected to each other or all not connected? • Input Format: • Line 1: N, M • Line 2...M+1: a, b: edge from a to b (1 <= a,b <= N) • Sample: N = 6, M = 3 • Edges: 1 2, 1 3, 2 3 • Output: 2 (123 are all connected, 456 are all not) • Constraints: 1 <= N,M <= 100000 • Partial Credit if it passes for small cases.

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