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)

object Main extends App {
    val scan = new java.util.Scanner(System.in)
    val n = scan.next().toInt
    
    val graph: Dijkstra.Graph = Array.fill(n)(List())

    for (_ <- 1 to n) {
      val u, k = scan.next().toInt
      graph(u) = (1 to k).toList
                .map(_ => Dijkstra.Edge(scan.next().toInt, scan.next().toInt))
    }

    Dijkstra.shortest_path(graph, 0,n-1).zip(0 until n)
      .map(t => t._2.toString ++ " " ++ t._1.toString)
      .foreach(println(_))
}

object Dijkstra {
    case class Edge(to: Int, cost: Long)
    type Edges = List[Edge]
    type Graph = Array[Edges]
    val inf = (1e9+7).toLong
    
    def shortest_path(graph: Graph, start: Int, goal: Int): Array[Long] = {
        val pq = new collection.mutable.PriorityQueue[(Long, Int)]
        val cost = Array.fill(graph.length)(inf)
        cost(start) = 0
        pq += ((cost(start), start))
        while (pq.nonEmpty) {
            val current_edge = pq.dequeue()
            if (cost(current_edge._2) < current_edge._1) {
            } else {
                for (tmp <- graph(current_edge._2)) {
                    val sumCost = current_edge._1 + tmp.cost
                    if (cost(tmp.to) > sumCost) {
                        cost(tmp.to) = sumCost
                        pq += ((cost(tmp.to), tmp.to))
                    }
                }
            }
        }
        cost
    }
}