Problem:
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Minimum Spanning Tree

For a given weighted graph $G = (V, E)$, find the minimum spanning tree (MST) of $G$ and print total weight of edges belong to the MST.

Input

In the first line, an integer $n$ denoting the number of vertices in $G$ is given. In the following $n$ lines, a $n \times n$ adjacency matrix $A$ which represents $G$ is given.
$a_{ij}$ denotes the weight of edge connecting vertex $i$ and vertex $j$. If there is no edge between $i$ and $j$, $a_{ij}$ is given by -1.

Output

Print the total weight of the minimum spanning tree of $G$.

Constraints

$1 \leq n \leq 100$

$0 \leq a_{ij} \leq 2,000$ (if $a_{ij} \neq -1$)

$a_{ij} = a_{ji}$

$G$ is a connected graph

Sample Input 1

5
-1 2 3 1 -1
2 -1 -1 4 -1
3 -1 -1 1 1
1 4 1 -1 3
-1 -1 1 3 -1

Sample Output 1

5

Reference

Introduction to Algorithms, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. The MIT Press.

Solution in Ruby (judge verdict: Accepted)

class DisjointSet
  def initialize(size = nil)
    @rank = Array.new(size){0}
    @parent = []
    if size
      size.times do |i|
        @parent[i] = i
      end
    end
  end

  def same(x, y)
    return find(x) == find(y)
  end

  def unite(x, y)
    x = find(x)
    y = find(y)
    if @rank[x] > @rank[y]
      @parent[y] = x
    else
      @parent[x] = y
      @rank[y] += 1 if @rank[x] == @rank[y]
    end
  end

  def find(x)
    @parent[x] = find(@parent[x]) unless x == @parent[x] 
    return @parent[x]
  end
end

n = gets.to_i
tree = DisjointSet.new(n)

sides = {}
n.times{|i|
  arr = gets.split.map(&:to_i)
  arr.each_with_index{|val, j|
    if i > j && val > -1
      sides["#{i} #{j}"] = val
    end
  }
}

sides = sides.sort_by {|k, v| v}
sum = 0
sides.each{|side|
  r, l = side[0].split.map(&:to_i)
  unless tree.same(r, l) then
    sum += side[1]
    tree.unite(r, l)
  end
}
puts sum