Problem:
Bridge Removal

ICPC islands once had been a popular tourist destination.
For nature preservation, however, the government decided to prohibit entrance to the islands,
and to remove all the man-made structures there.
The hardest part of the project is to remove all the bridges connecting the islands.

There are n islands and n-1 bridges.
The bridges are built so that all the islands are reachable from all the other islands by
crossing one or more bridges.
The bridge removal team can choose any island as the starting point, and can repeat either of the
following steps.

Move to another island by crossing a bridge that is connected to the current island.

Remove one bridge that is connected to the current island, and stay at the same island after the removal.

Of course, a bridge, once removed, cannot be crossed in either direction.
Crossing or removing a bridge both takes time proportional to the length of the bridge.
Your task is to compute the shortest time necessary for removing all the bridges.
Note that the island where the team starts can differ from where the team finishes the work.

Input

The input consists of at most 100 datasets.
Each dataset is formatted as follows.

n

p2 p3 ... pn

d2 d3 ... dn

The first integer n (3 ≤ n ≤ 800) is the number of the islands.
The islands are numbered from 1 to n.
The second line contains n-1 island numbers pi
(1 ≤ pi < i), and tells that for each i
from 2 to n the island i and the island pi are connected by a bridge.
The third line contains n-1 integers di (1 ≤ di ≤ 100,000)
each denoting the length of the corresponding bridge.
That is, the length of the bridge connecting the island i and pi
is di. It takes di units of time to cross the bridge, and
also the same units of time to remove it.

Note that, with this input format, it is assured that all the islands are reachable each other by
crossing one or more bridges.

The input ends with a line with a single zero.

Output

For each dataset, print the minimum time units required to remove all the bridges in a single line.
Each line should not have any character other than this number.

Sample Input

4
1 2 3
10 20 30
10
1 2 2 1 5 5 1 8 8
10 1 1 20 1 1 30 1 1
3
1 1
1 1
0

Output for the Sample Input

80
136
2

Solution in Ruby (judge verdict: Accepted)

class Heap
	require 'forwardable'
	extend Forwardable

	def initialize
		@array = []
		@index = {}
	end

	def add(obj, key)
		@array << [obj, key]
		@index[obj] = @array.size - 1
		lift_up(@array.size - 1)
	end

	def swap(i, j)
		u = @array[i]
		v = @array[j]
		@index[u[0]], @index[v[0]] = j, i
		@array[i], @array[j] = v, u
	end

	def lift_up(i)
		return if i == 0
		j = (i - 1) / 2
		if (@array[i][1] <=> @array[j][1]) < 0
			swap(i, j)
			lift_up(j)
		end
	end

	def del_min
		raise "heap is empty!" if empty?
		min = @array[0]
		if size > 1
			swap(0, @array.size-1)
			@array.pop
			lift_down(0)
		else
			@array.pop
		end
		@index.delete(min[0])
		min
	end

	def inc_key(obj, diff)
		i = @index[obj]
		@array[i][1] += diff
		lift_down(i)
	end

	def dec_key(obj, diff)
		i = @index[obj]
		@array[i][1] -= diff
		lift_up(i)
	end

	def update_key(obj, new_key)
		i = @index[obj]
		return unless i
		return if (new_key <=> @array[i][1]) >= 0
		@array[i][1] = new_key
		lift_up(i)
	end

	def min_obj
		@array[0][0]
	end

	def min_key
		@array[0][1]
	end

	def lift_down(i)
		j = 2 * i + 1
		k = j + 1
		if k >= size
			return if j >= size
			if (@array[i][1] <=> @array[j][1]) > 0
				swap(i, j)
				lift_down(j)
			end
		else
			l = ((@array[j][1] <=> @array[k][1]) < 0 ? j : k)
			if (@array[i][1] <=> @array[l][1]) > 0
				swap(i, l)
				lift_down(l)
			end
		end
	end

	def_delegators :@array, :empty?, :size
end

module Graphical
	require 'set'

	def dijkstra(start)
		raise "start must be Enumerable" if !start.kind_of?(Enumerable)
		h = Heap.new
		reached = Set[*start.to_a]
		start.each{|s| h.add(s, 0)}

		dist = {}
		until h.empty?
			u, key = h.del_min
			return key if block_given? && yield(u, key)
			dist[u] = key
			each_out_connection_of(u){|v, l|
				d = dist[u] + l
				if reached.include?(v)
					h.update_key(v, d)
				else
					h.add(v, d)
					reached << v
				end
			}
		end
		block_given? ? nil : dist
	end
end

class Graph
	include Graphical

	def initialize(*options)
		@options = Set[*options]
	end

	def add_node(property = {})
		Node.new(property)
	end

	def add_edge(u, v, property = {})
		e = Edge.new(u, v, property)
		u.add_edge(e)
		v.add_edge(e)
		e
	end
	
	def del_edge(e)
		u, v = e.opposite.keys
		u.del_edge(e)
		v.del_edge(e)
		e
	end

	class Node
		def initialize(property)
			@property = property
			@edges = Set.new
		end

		def add_edge(e)
			@edges << e
		end

		def del_edge(e)
			@edges.delete(e)
		end

		def to_s
			@property[:name] || super
		end

		def each_edge
			@edges.each{|e| 
				yield e
			}
		end

		attr_reader :property, :edges
	end

	class Edge
		def initialize(u, v, property)
			@opposite = {u => v, v => u}
			@property = property
		end
		attr_reader :opposite, :property

		def to_s
			@property[:name] || super
		end

		def length
			@property[:length] || 1
		end
	end

	def each_out_connection_of(u)
		u.each_edge{|e|
			yield(e.opposite[u], e.length)
		}
	end

	alias :each_in_connection_of :each_out_connection_of
end

loop do
	n = gets.to_i
	break if n == 0
	g = Graph.new
	nodes = (0..n-1).map { g.add_node }

	link = gets.split.map(&:to_i)
	len = gets.split.map(&:to_i)
	all_len = len.inject(:+)
	(0..n-2).each do |i|
		g.add_edge(nodes[i+1], nodes[link[i]-1], {:length => len[i]})
	end

	leaf_len = 0
	to_be_deleted = []
	nodes.each do |v|
		edges = v.edges
		if edges.size == 1 
			leaf_len += edges.to_a[0].property[:length]
			to_be_deleted << v
		end
	end

	to_be_deleted.each do |v|
		nodes.delete(v)
		e = v.edges.to_a[0]
		e.opposite[v].del_edge(e)
	end

	dist = g.dijkstra([nodes[0]])
	u = dist.keys.max_by{|u| dist[u]}
	diameter = g.dijkstra([u]).values.max

	p all_len * 3 - leaf_len * 2 - diameter
end