USACO Training: Cow Tours (and Floyd-Warshall's)

-

USACO Training: Cow Tours

Idea:

Let us first initially use Floyd-Warshall's to calculate the all-pairs shortest paths.
Afterwords, our first instinct would be to do the following:
  1. Select a node i,
  2. Select a node j,
  3. Create an edge between i and j,
  4. Do FW again,
  5. Loop through array to find the maximum edge.
However, the problem is soon obvious: this runs in O(n&5) time: o(n) for selecting i, o(n)for selecting j, o(n^3) for doing the FW, and O(n&3) for looping though the array.

How can we optimize this?

Let us introduce a new measure, the longest possible (non-infinite) distance between a node X and any other node. Let it be stored in an array called "far". We notice that since we have already done Floyd-Warshalls on the grid, we can precalculate this array. Notice that now, we instantly reduce the time by a factor of O(n^4) - we can instantly, i.e. in O(1), calculate the diameter if we added an edge between I and J!

However, we would need to calculate the diameters first.
How?
The diameter of a node X is the collective maximum of far[i] as long as dist[X][i] is not infinity.

The diameter if we collect node I and node J is the maximum of the following;
  1. diameter of I,
  2. diameter of J,
  3. far[i] + far[j] + EuclideanDistance(i, j)

Bug List

1. I hard-coded a permanent 2 fields into my program. However, it isn't important - all we need are to connect any two fields.
2. Floyd-Warshalls:

Floyd-Warshall Loop Order

K, I, J

Explanation -
If it were I, J, K, if K acted as an intermediate point, the path I -> J would only get one intermediate point.
However, if we looped over K as a set of intermediate points, then I -> J would get all the intermediate points that it needed.

If the loop order is I, J, K, we would keep K as the intermediate point. However, choosing the maximum would ALWAYS result in only 1 intermediate point being used.

 

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

UVA 787 and Big Number Multiplication

-
UVA 787 and Big Number Multiplication Problem Background: Big Numbers in this problem Big numbers are effectively storing sequence of digit chunks and directly manipulating them as if they were a large number. The most intuitive way to do this is to make an array where every element is a digit, but this is not effectively using the memory. Hence, it is logical to store numbers "millions at a time", i.e. 324876498726832746  [2]  | [1] | [0] like that. Necessities to Implement Notice that although this problem may LOOK like DP, it's really solvable by O(N^2) enumeration. Also notice that in this enumeration we will only be multiplying a large number with a normal number. Pseudocode for algorithm: CREATE LargeNumber NAME best SET -INFINITY FOR EACH i BETWEEN 0 and LENGTH OF list DO:     CREATE LargeNumber NAME product SET 1     FOR EACH j BETWEEN i AND LENGTH OF list DO         SET product TO product TIMES list[j] ...
Read more