USACO 2018 Open: Milking Order and Topological Sort

-


USACO 2018 US Open Contest, Gold

Problem 2. Milking Order


We need to find the biggest number X such that the first X observations are satisfied.
Notice that this is perfect to do a binary search on: the invariant is that if the first X observations are satisfied, all observations X-1 will be satisfied; if the first X are insatisfied, all observations X+1 will not be satisfied.
So how do we check if all observations to X are satisfied?
Let us assemble a graph, in which each observation O_1, O_2, O_3, ... O_i, ... O_l(o) is represented as an edge between O_i and O_i+1 for all i. These observations are satisfiable if and only if this graph has a topological sorting.
Lexicographic topological sorting on a graph can be done in O(E + V log_2 V) time (log_2 V because of priority_queue to extract lowest element). So, our total time complexity will be O((E + NlogN) logM). Here, E=200K (the sum of the M_i), N=100K, and M=50K, so it's fast enough.
LTS can be easily implemented with a priority queue:

  • Assemble the graph. Whilst adding edges, keep track of how many directed edges point towards a node.
  • Add every node that isn't pointed to into the priority queue.
  • (BFS-ish part) While the priority queue is not empty, pop the lowest element.
  • Add this element to the LTS sequence.
  • Remove all edges extending from that element, and update the in-edge count accordingly.
  • If there are any new nodes with an in-edge count of 0, push it into the queue.
  • Continue until the sequence is empty.
  • If any nodes were not visited, LTS is impossible.
  • Else the LTS is in the sequence.

Comments

Post a Comment

Popular posts from this blog

USACO Training "subset": Subset Sums

-
USACO Training "subset": Subset Sums Problem Statement Subset Sums JRM For many sets of consecutive integers from 1 through N (1 <= N <= 39), one can partition the set into two sets whose sums are identical. For example, if N=3, one can partition the set {1, 2, 3} in one way so that the sums of both subsets are identical: {3} and {1,2} This counts as a single partitioning (i.e., reversing the order counts as the same partitioning and thus does not increase the count of partitions). If N=7, there are four ways to partition the set {1, 2, 3, ... 7} so that each partition has the same sum: {1,6,7} and {2,3,4,5} {2,5,7} and {1,3,4,6} {3,4,7} and {1,2,5,6} {1,2,4,7} and {3,5,6} Given N, your program should print the number of ways a set containing the integers from 1 through N can be partitioned into two sets whose sums are identical. Print 0 if there are no such ways. Your program must calculate the answer, not look it up from a table. PROGRAM NAME: s
Read more

USACO 2018 Open: Talent Show

-
Image
USACO 2018 Open Talent Show In Short Given N objects with value Vi and weight Wi, a collection of elements is said to have total value-to-weight ratio as (the sum of their value) divided by (the sum of their weight). Find the best possible value-to-weight ratio of these X objects, given the restraint that the collection must have total weight at least W. Return that ratio multiplied by 1000 and floored. Analysis It would be very difficult to calculate this optimal ratio - we'd have to either try all possible collections - O(2^n), where n is at most 250, about 10 to the power of 75 calculations, OR use a greedy heuristic - get the objects with the best ratio until it satisfies the weight limit. Neither of those are too to our liking, so how about searching for it? The search space is linear and not infinite - it asks for the ratio to the closest thousandths. The search space is sorted as well - if ratio X is achievable, then ratio Y < X is also achievable, if
Read more

OpenJudge 1768: Kadane's Algorithm on 2D Arrays

-
OpenJudge 1768: Kadane's Algorithm on 2-D Arrays Simple Version Given a 2-D matrix of integers, find the sub-matrix of this whose sum is maximized, and find that sum. The size of the matrix is no larger than 100 by 100. Analysis O(N^4) - Works, but naive     That's true, an naive O(N^4) version would work. We simply need to calculate (two dimensional!) prefix sums first. A quick review to what 2-D prefix sums are:     The 2-D prefix sum of two coordinates, X and Y, is the sum of all elements that have X-coordinate less than or equal to X, and Y coordinate less than or equal to Y.     It can be calculated through tabulation by removing overlapping areas.     Let P be the prefix sum array and V be the array of values, so the formula is as follows:     P[x, y] = V[x, y] - P[x-1, y] - P[x, y-1] + P[x-1, y-1].     We need to add the P[x-1, y-1] at the end since subtracting P[x-1, y] and P[x, y-1] we had overlapped the P[x-1, y-1]. Then, all we need to
Read more