Problem:
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Single Source Shortest Path

For a given weighted graph $G = (V, E)$, find the shortest path from a source to each vertex. For each vertex $u$, print the total weight of edges on the shortest path from vertex $0$ to $u$.

Input

In the first line, an integer $n$ denoting the number of vertices in $G$ is given. In the following $n$ lines, adjacency lists for each vertex $u$ are respectively given in the following format:

$u$ $k$ $v_1$ $c_1$ $v_2$ $c_2$ ... $v_k$ $c_k$

Vertices in $G$ are named with IDs $0, 1, ..., n-1$. $u$ is ID of the target vertex and $k$ denotes its degree. $v_i (i = 1, 2, ... k)$ denote IDs of vertices adjacent to $u$ and $c_i$ denotes the weight of a directed edge connecting $u$ and $v_i$ (from $u$ to $v_i$).

Output

For each vertex, print its ID and the distance separated by a space character in a line respectively. Print in order of vertex IDs.

Constraints

$1 \leq n \leq 100$

$0 \leq c_i \leq 100,000$

$|E| \leq 10,000$

All vertices are reachable from vertex $0$

Sample Input 1

5
0 3 2 3 3 1 1 2
1 2 0 2 3 4
2 3 0 3 3 1 4 1
3 4 2 1 0 1 1 4 4 3
4 2 2 1 3 3

Sample Output 1

0 0
1 2
2 2
3 1
4 3

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.io.StdIn

object Main extends App {

  val n = StdIn.readLine().toInt
  val M = Array.fill[Int](n, n)(Int.MaxValue)

  (0 until n).foreach { _ =>
    val a = StdIn.readLine().split(' ').map(_.toInt)
    (0 until a(1)).foreach { k =>
      M(a(0))(a(2 + k * 2)) = a(2 + k * 2 + 1)
    }
  }

  // (0 until n).foreach { i => println(M(i).mkString("")) }
  private val WHITE = 0
  private val GRAY = 1
  private val BLACK = 9

  val color = Array.fill[Int](n)(WHITE)
  val distance = Array.fill(n)(Int.MaxValue)
  val parent = Array.fill(n)(-1)

  def dijkstra(s: Int): Unit = {
    var mincost = -1
    var u = 0
    distance(s) = 0
    parent(s) = -1

    while (mincost != Int.MaxValue) {
      mincost = Int.MaxValue
      (0 until n).foreach { i =>
        if (color(i) != BLACK && distance(i) < mincost) {
          mincost = distance(i)
          u = i
        }
      }

      if (mincost != Int.MaxValue) {
        color(u) = BLACK

        (0 until n).foreach { v =>
          if (color(v) != BLACK && M(u)(v) != Int.MaxValue) {
            if (distance(u) + M(u)(v) < distance(v)) {
              distance(v) = distance(u) + M(u)(v)
              parent(v) = u
              color(v) = GRAY
            }
          }
        }

      }

    }
  }

  dijkstra(0)

  (0 until n).foreach(i => println(i + " " + distance(i)))

}