Solution to Day 12: Digital Plumber - Part 1

advent_of_code_d12_p1.py

#aoc_d12.py
import re

inp = open('input_d12.txt','r').readlines()
#inp = ["0 <-> 2","1 <-> 1", "2 <-> 0, 3, 4", "3 <-> 2, 4", "4 <-> 2, 3, 6", "5 <-> 6", "6 <-> 4, 5"]

count = 0
connections = {}
visited = []

def check(items):
	#print items
	if 0 == items:
		#print 'yup'
		return True
	if type(items) is int and items not in visited:
		#print 'loop for ',items,visited
		visited.append(items)
		r = check(connections[items])
		#visited.remove(items)
		#print 'loop result:',r
		return r
	elif type(items) is int and items in visited:
		#print '=> f', visited
		return False
	
	#print 'moar check'
	for i in items:
		#print 'loop for ',i,visited
		if check(i):
			return True
	#print '=> f2'
	return False

for i in inp:
	if "," in i:
		m = re.match("(\d+)\s<->\s([0-9,\s]+)",i)
		#print i
		i1 = int(m.group(1))
		#print m.group(2)
		items = [int(e) for e in m.group(2).split(',')]
		connections[i1] = items
		#for it in items:
		#	connections[it] = i1
	if "," not in i:
		m = re.match("(\d+)\s<->\s(\d+)",i)
		i1 = int(m.group(1))
		i2 = int(m.group(2))
		if i1 in connections:
			connections[i1].append(i2)
		else:
			connections[i1] = [i2]
		if i2 in connections:
			connections[i2].append(i1)
		else:
			connections[i2] = [i1]
#print connections

for k in connections:
	visited = []
	if connections[k] is int and connections[k] == 0:
		#print "+1 for k=",k
		count += 1
	else:
		if check(connections[k]):
			#print "+1 for k=",k
			count += 1
		else:
			pass#print "no for k=",k
	#print '-----'
print count