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)

def setup_edges(matrix)
  edges = []
  matrix.each_with_index do |row, ind_row|
    row.each_with_index do |col, ind_col|
      if col >= 0 then
        edges << [ind_col+1, ind_row+1, col]
      end
    end
  end
  return edges.sort_by {|edge| edge[2]}
end


def solve(edges)
  sum = 0
  vertices = Hash.new
  (edges.size).times do |i|
    vertices[i] = i
  end

  until edges.empty? do
    edge = edges.shift
    v1, v2 = vertices[edge[0]-1], vertices[edge[1]-1]
    next if v1 == v2
    sum += edge[2]
    min_id = [v1, v2].min
    vertices.each do |key, val|
      vertices[key] = min_id if val == v1 || val == v2
    end
  end
  return sum
end


# main
n = gets.chomp.to_i
matrix = []
(n).times do 
  matrix << gets.chomp.split.map(&:to_i)
end

edges = setup_edges(matrix)
puts solve(edges)