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Problem solution · Python

CCC 2011 S4 - Blood Distribution

CCC 2011 S4 - Blood Distribution: a Python solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to CCCSolutions.

Technique
Direct simulation
Source
CCCSolutions
Length
129 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

Direct simulation

For CCC 2011 S4 - Blood Distribution, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 129 lines of Python from the credited upstream file ccc11s4.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

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 2011 S4 - Blood Distribution · PythonPython
Use this to learn the idea, then write your own version.
# CCC 2011 Senior 4: Blood Distribution# written by C. Robart# March 2011 # the order in which you match up the types makes little# difference in the CCc test cases# However Kevin Luo pointed out that problems can arise with# certain orders, so two different ones are tried and the max taken## eg:#    0 0 1 0 1 0 0 0#    0 0 0 0 0 1 1 0#    #    0 0 1 0 1 0 0 0#    0 0 0 1 0 0 1 0#    #    0 1 1 0 0 0 0 0#    0 0 0 1 0 1 0 0#    #    0 1 1 0 0 0 0 0#    0 0 0 1 0 0 1 0#    #    0 1 0 0 1 0 0 0#    0 0 0 1 0 1 0 0#    #    0 1 0 0 1 0 0 0#    0 0 0 0 0 1 1 0##    All the above have an answer of 2. In single ordering will not do that. ##    0 0 2 0 2 0 0 0#    0 0 0 1 0 1 2 0##    Here is one with an answer of 4.   #finds the min amount of blood[b] and patients[p]#returns it and reduces the two lists (arrays)def units(b,p):    u = min(blood[b], patients[p])    blood[b] = blood[b] - u    patients[p] = patients[p] - u     return u  # file inputfile = open("s4_1_in.txt", 'r')line = file.readline()blood = []for i in range (0,7):    space = line.find(" ")    blood.append( eval(line[0:space]))    line = line[space+1:]blood.append(eval(line))     line = file.readline()patients = []for i in range (0,7):    space = line.find(" ")    patients.append( eval(line[0:space]))    line = line[space+1:]patients.append(eval(line))     holdpatients = list(patients)holdblood = list(blood) # try this orderingtotal = 0# O- (take all O- you can)total = total + units(0,0)# O+ (take all O+ you can, then take from 0-)total = total + units(1,1) + units(0,1)# A- (take all A- you can, then take from 0-)total = total + units(2,2) + units(0,2)# B- (take all B- you can, then take from 0-)total = total + units(4,4) + units(0,4) # A+ (take all A+ you can, then take from O+)total = total + units(3,3) + units(1,3) # B+ (take all B+ you can, then take from O+)total = total + units(5,5) + units(1,5)# A+ (take all A- you can, then take from O-)total = total + units(2,3) + units(0,3) # B+ (take all B- you can, then take from O-)total = total + units(4,5) + units(0,5) # AB- (take all AB- you can, then take from B-, A- or O-)total = total + units(6,6) + units(4,6) + units(2,6) + units(0,6)# AB+ (take all AB+ you can, then take from AB-, B+, B-, A+, A-, O+ or O-)total = total + units(7,7) + units(6,7) + units(5,7) + units(4,7) + units(3,7) + units(2,7) + units(1,7) + units(0,7) print "first order",total patients = holdpatientsblood = holdblood # try this second orderingtotal2 = 0# O- (take all O- you can)total2 = total2 + units(0,0)# O+ (take all O+ you can, then take from 0-)total2 = total2 + units(1,1) + units(0,1)# A- (take all A- you can, then take from 0-)total2 = total2 + units(2,2) + units(0,2)# B- (take all B- you can, then take from 0-)total2 = total2 + units(4,4) + units(0,4) # A+ (take all A+ you can, then take from A-)total2 = total2 + units(3,3) + units(2,3)# B+ (take all B+ you can, then take from B-)total2 = total2 + units(5,5) + units(4,5)# A+ (take all O+ you can, then take from O-)total2 = total2 + units(1,3) + units(0,3) # B+ (take all O+ you can, then take from O-)total2 = total2 + units(1,5) + units(0,5) # AB- (take all AB- you can, then take from B-, A- or O-)total2 = total2 + units(6,6) + units(4,6) + units(2,6) + units(0,6)# AB+ (take all AB+ you can, then take from AB-, B+, B-, A+, A-, O+ or O-)total2 = total2 + units(7,7) + units(6,7) + units(5,7) + units(4,7) + units(3,7) + units(2,7) + units(1,7) + units(0,7) print "second order",total2  print max (total, total2)

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