Rubyでdijkstra法

Rubyでダイクストラ法を使って始点からの最短経路を求めることができるかもしれないコード

nodes = ["A", "B", "C", "D", "E"]
connections = [
	# [node1, node2, cost(1->2), cost(2->1)]
	["A", "B", 6, 9],
	["A", "C", 3, 2],
	["B", "C", 2, 3],
	["B", "D", 4, 1],
	["C", "E", 3, 8],
	["B", "E", 3, 2],
]
require 'pp'

def dijkstra(nodes, connections, start)
	inf = 100000
	d = {}
	cost = {}
	used = {}
	prev = {}
	nodes.each{|n|
		d[n] = {}
		nodes.each{|m|
			d[n][m] = inf
		}
		cost[n] = inf
		used[n] = 0
	}
	connections.each{|c|
		d[c[1]][c[0]] = c[2] ? c[2] : 1
		d[c[0]][c[1]] = c[3] ? c[3] : (c[2]?c[2]:1)
	}
	cost[start] = 0
	while true
		min_cost = inf
		nodes.each{|node|
			if used[node] == 0 && min_cost > cost[node]
				min_cost = cost[node]
			end
		}
		break if min_cost == inf
		nodes.each{|src|
			next if min_cost != cost[src]
			nodes.each{|dst|
				next if cost[dst] <= d[dst][src] + cost[src]
				cost[dst] = d[dst][src] + cost[src]
				prev[dst] = src
			}
			used[src] = 1
		}
	end
	routes = {}
	nodes.each{|dst|
		route = []
		node = dst
		while node != nil && node != start
			route << node
			node = prev[node]
		end
		routes[dst] = route.reverse
	}
	return routes
end

routes = {}
nodes.each{|st|
	rs = dijkstra(nodes, connections, st)
	rs.each{|d,r|
		k = st+'-'+d
		routes[k] = []
		m = st
		r.each{|n|
			routes[k] << m+'-'+n
			m = n
		}
	}
}

pp routes

{"B-A"=>["B-C", "C-A"],
 "A-B"=>["A-B"],
 "C-A"=>["C-A"],
 "B-B"=>[],
 "A-C"=>["A-C"],
 "D-A"=>["D-B", "B-C", "C-A"],
 "C-B"=>["C-B"],
 "B-C"=>["B-C"],
 "A-D"=>["A-B", "B-D"],
 "E-A"=>["E-B", "B-C", "C-A"],
 "D-B"=>["D-B"],
 "C-C"=>[],
 "B-D"=>["B-D"],
 "A-E"=>["A-C", "C-E"],
 "E-B"=>["E-B"],
 "D-C"=>["D-B", "B-C"],
 "C-D"=>["C-B", "B-D"],
 "B-E"=>["B-E"],
 "E-C"=>["E-B", "B-C"],
 "D-D"=>[],
 "C-E"=>["C-E"],
 "E-D"=>["E-B", "B-D"],
 "D-E"=>["D-B", "B-E"],
 "E-E"=>[],
 "A-A"=>[]}