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: Wrong Answer)

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

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

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

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

  val parents = mutable.Seq.fill(n)(-1)
  val distances = mutable.Seq.fill(n)(Int.MaxValue)
  distances(0) = 0
  val que = mutable.PriorityQueue(Edge(0, 0))(Ordering.by((e: Edge) => e.cost).reverse)

  while (que.nonEmpty) {
    val edge = que.dequeue()
    if (distances(edge.to) >= edge.cost) {
      for (e <- d(edge.to)) {
        if (e.cost != -1 &&
          distances(e.to) > distances(edge.to) + e.cost) {
          distances(e.to) = distances(edge.to) + e.cost
          parents(e.to) = edge.to
          que.enqueue(e)
        }
      }
    }
  }

  var sum = 0
  for (i <- parents.tail.indices) {
    sum += d(parents(i + 1))(i + 1).cost
  }

  println(sum)
}