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)

INF = 0x7fffffff

def prim(g, n)
  d = Array.new(n, INF)
  pi = Array.new(n, -1)
  color = Array.new(n, 0)
  
  d[0] = 0  
  sum = 0
  loop do
    u = -1
    mincost = INF
    (0...n).each do |i|
      if mincost > d[i] && color[i] != 2
        u = i
        mincost  = d[i]
      end
    end
    break if u == -1
    color[u] = 2
    
    (0...n).each do |i|
      if color[i] != 2 && d[i] > g[u][i] && g[u][i] != INF
          d[i] = g[u][i]
          pi[i] = u
          color[i] = 1
      end
    end
  end
  (0...n).each do |j|
    sum += g[j][pi[j]] unless pi[j] == -1
  end
  return sum
end

n = gets.to_i
g = Array.new(n) { Array.new(n) }

(0...n).each do |i|
  g[i] = gets.split.map(&:to_i)
  (0...n).each do |j|
    g[i][j] = INF if g[i][j] == -1
  end
end
puts prim(g, n)