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.{readLine,readInt}
import scala.collection.mutable.ArrayBuffer
 
object Main {
  def main(args:Array[String]) = {
    val n = readInt
    val g = for(i<-0 until n) yield { readLine.trim.split(" ").map(_.toInt) }
    val dist = new Array[Int](n)
    val visited = new ArrayBuffer[Int]
    for(i<-0 until n) dist(i) = Int.MaxValue
    dist(0) = 0
 
    while(visited.length < n) {
      val minV = (for(i<-0 until dist.length if !visited.contains(i)) yield (dist(i),i)).min
      val min = minV._1
      val v = g(minV._2)
 
      for(i<-0 until v(1) if !visited.contains(v(i*2+2))) {
        if(dist(v(i*2+2)) == Int.MaxValue) dist(v(i*2+2)) = v(i*2+3) + dist(v(0))
        else if(v(i*2+3)+dist(v(0)) < dist(v(i*2+2))) dist(v(i*2+2)) = v(i*2+3)+dist(v(0))
      }
      visited += v(0)
    }
    for(i<-0 until n) {
      println(i + " " + dist(i))
    }
  }
}