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Problem solution · Python

CCC 2014 S5 - Lazy Fox

CCC 2014 S5 - Lazy Fox: a Python solution using sorting and greedy selection. Learn the idea, check the complexity, and read the full code, with credit to CCCSolutions.

Technique
Sorting and greedy selection
Source
CCCSolutions
Length
126 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Sorting and greedy selection

For CCC 2014 S5 - Lazy Fox, the implementation first exposes a useful order, then scans that order while making locally justified choices.

  1. Choose the key that reveals the greedy or grouping structure.
  2. Sort the relevant records by that key.
  3. Scan in order, maintaining the invariant that makes each local choice safe.

Code notes

  • 126 lines of Python from the credited upstream file ccc14s5.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Sorting is typically the dominant term unless the subsequent scan uses a more expensive nested operation.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from CCCSolutions by CCCSolutions contributors · Milliken Mills High School and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCCC 2014 S5 - Lazy Fox · PythonPython
Use this to learn the idea, then write your own version.
# CCC 2014 Senior 5: Lazy Fox## This solution is by Jacob Jackson of University of Toronto Schools## This is Jackson's explaination of the code:## Make a list of all points and add the origin to that list.# Take all pairs of points and sort them in increasing order of distance.# Process these pairs in order, keeping an array best[N],# which represents the maximum number of treats attainable# from a given location if you can't travel farther than the current distance# (the current distance = the distance between the two points# you're currently processing).# For each pair (A, B), consider traveling from A to B or from B to A. That is,#     best[A] = max(best[A], best[B] + 1);#     best[B] = max(best[B], best[A] + 1);## Here is a trace of the variables as it processes the example data set# given with the problem statement:##   pairs [[1, 3, 4], [1, 4, 5], [4, 3, 5], [5, 1, 2], [10, 0, 3],#          [13, 0, 4], [18, 0, 5], [29, 1, 5], [40, 1, 4], [50, 2, 5],#          [53, 1, 3], [65, 2, 4], [82, 2, 3], [89, 0, 1], [116, 0, 2]]##   pair= [1, 3, 4]#   pdist= [0, 0, 0, 1, 1, 0]   pbest= [0, 0, 0, 0, 0, 0]   best= [0, 0, 0, 1, 1, 0]#   #   pair= [1, 4, 5]#   pdist= [0, 0, 0, 1, 1, 1]   pbest= [0, 0, 0, 0, 0, 0]   best= [0, 0, 0, 1, 1, 1]#   #   pair= [4, 3, 5]#   pdist= [0, 0, 0, 4, 1, 4]   pbest= [0, 0, 0, 1, 0, 1]   best= [0, 0, 0, 2, 1, 2]#   #   pair= [5, 1, 2]#   pdist= [0, 5, 5, 4, 1, 4]   pbest= [0, 0, 0, 1, 0, 1]   best= [0, 1, 1, 2, 1, 2]#   #   pair= [10, 0, 3]#   pdist= [10, 5, 5, 10, 1, 4]   pbest= [0, 0, 0, 2, 0, 1]   best= [2, 1, 1, 2, 1, 2]#   #   pair= [13, 0, 4]#   pdist= [13, 5, 5, 10, 13, 4]   pbest= [2, 0, 0, 2, 1, 1]   best= [2, 1, 1, 2, 1, 2]#   #   pair= [18, 0, 5]#   pdist= [18, 5, 5, 10, 13, 18]   pbest= [2, 0, 0, 2, 1, 2]   best= [2, 1, 1, 2, 1, 2]#   #   pair= [29, 1, 5]#   pdist= [18, 29, 5, 10, 13, 29]   pbest= [2, 1, 0, 2, 1, 2]   best= [2, 3, 1, 2, 1, 2]#   #   pair= [40, 1, 4]#   pdist= [18, 40, 5, 10, 40, 29]   pbest= [2, 3, 0, 2, 1, 2]   best= [2, 3, 1, 2, 4, 2]#   #   pair= [50, 2, 5]#   pdist= [18, 40, 50, 10, 40, 50]   pbest= [2, 3, 1, 2, 1, 2]   best= [2, 3, 3, 2, 4, 2]#   #   pair= [53, 1, 3]#   pdist= [18, 53, 50, 53, 40, 50]   pbest= [2, 3, 1, 2, 1, 2]   best= [2, 3, 3, 4, 4, 2]#   #   pair= [65, 2, 4]#   pdist= [18, 53, 65, 53, 65, 50]   pbest= [2, 3, 3, 2, 4, 2]   best= [2, 3, 5, 4, 4, 2]#   #   pair= [82, 2, 3]#   pdist= [18, 53, 82, 82, 65, 50]   pbest= [2, 3, 5, 4, 4, 2]   best= [2, 3, 5, 6, 4, 2]#   #   pair= [89, 0, 1]#   pdist= [89, 89, 82, 82, 65, 50]   pbest= [2, 3, 5, 4, 4, 2]   best= [3, 3, 5, 6, 4, 2]#   #   pair= [116, 0, 2]#   pdist= [116, 89, 116, 82, 65, 50]   pbest= [3, 3, 5, 4, 4, 2]   best= [5, 3, 5, 6, 4, 2]### Times of the test data sets on my Pentium 4, 3.2Ghz processor.#     1 to 9 all well less than 1 sec#     10 = 5.5 sec#     11 = 25 sec#     12 = 16 sec#     13 = 20 sec#     14 = 20 sec#     15 = 26 sec# Several other people sent me solutions, but# NONE were as fast as this. Not even close. So if you think you can# do better, run this on your machine as a benchmark and # write a faster PYTHON program. And no fair using anything# built-in other than sort() and normal array processing. :-)  file = open ("s5.15.in", "r")N = int(file.readline()) pts = [[0, 0]]for i in range(N):    pt = file.readline().split()    pts.append([int(pt[0]), int(pt[1])]) pairs = []for a in range(N+1):    for b in range(a+1, N+1):        dx = pts[a][0] - pts[b][0]        dy = pts[a][1] - pts[b][1]        pairs.append([dx * dx + dy * dy, a, b]) pairs.sort() best  = [0] * (N+1)pbest = [0] * (N+1)pdist = [0] * (N+1) for pair in pairs:    d = pair[0]    a = pair[1]    b = pair[2]     if d != pdist[a]:        pdist[a] = d        pbest[a] = best[a]    if d != pdist[b]:        pdist[b] = d        pbest[b] = best[b]     if a == 0: # the origin is a special case because we cannot revisit it        best[a] = max(best[a], pbest[b])    else:        best[a] = max(best[a], pbest[b] + 1)        best[b] = max(best[b], pbest[a] + 1)     print best[0] + 1

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