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 Scala (judge verdict: Accepted)

import scala.collection.mutable
import scala.io.StdIn._

case class Edge(from: Int, to: Int, cost: Int)

object Main extends App {
  val n = readInt()
  val d = mutable.Seq.fill(n, n)(Int.MaxValue)
  var edges = mutable.Seq.fill(n)(Seq.empty[Edge])

  for (i <- 0 until n) {
    val costs = readLine().tail.split(" ")
    for (j <- costs.indices) {
      if (costs(j) != "-1") {
        d(i)(j) = costs(j).toInt
        edges(i) :+= Edge(i, j, costs(j).toInt)
      }
    }
  }

  val searched = mutable.Seq.fill(n)(false)
  val parents = mutable.Seq.fill(n)(-1)
  val distances = mutable.Seq.fill(n)(Int.MaxValue)
  distances(0) = 0
  val que = mutable.PriorityQueue(0)(Ordering.by((x: Int) => distances(x)).reverse)

  while (que.nonEmpty) {
    val from = que.dequeue()
    if (!searched(from)) {
      searched(from) = true
      edges(from).foreach(e => {
        val to = e.to
        if (!searched(to) && distances(to) > d(from)(to)) {
          distances(to) = d(from)(to)
          parents(to) = from
        }
        que.enqueue(to)
      })
    }
  }

  val s = parents.zipWithIndex
    .foldLeft(0)((acc, z) =>
      z match {
        case (p, i) if p != -1 => acc + d(i)(p)
        case _ => acc
      })

  println(s)
}