Problem:
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Single Source Shortest Path II

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

$1 \leq n \leq 10,000$

$0 \leq c_i \leq 100,000$

$|E| < 500,000$

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 Ruby (judge verdict: Time Limit Exceeded)

class Priority_queue
  def initialize(bool = true)
    if bool
      @ope = proc{|a, b|(a and b)? a > b : false}
    else
      @ope = proc{|a, b|(a and b)? a < b : false}
    end
    @heap = []
  end

  def <<(other)
    @heap.push(other)
    up
  end

  def gets
    result = @heap.shift
    @heap.unshift(@heap.pop)
    down
    return result
  end
  
  def up(key = @heap.size - 1)
    if @ope.call(@heap[key / 2], @heap[key])
      @heap[key / 2], @heap[key] = @heap[key], @heap[key / 2]
      down(key / 2)
      up(key / 2)
    end
  end

  def down(key = 0)
    if @ope.call(@heap[key], @heap[key * 2]) and @ope.call(@heap[key * 2 + 1], @heap[key * 2])
      @heap[key], @heap[key * 2] = @heap[key * 2], @heap[key]
      down(key * 2)
    elsif @ope.call(@heap[key], @heap[key * 2 + 1])
      @heap[key], @heap[key * 2 + 1] = @heap[key * 2 + 1], @heap[key]
      down(key * 2 + 1)
    end
  end
  def inspect
    @heap
  end
end
class Node
  include Comparable
  attr_reader :cost, :id
  def initialize(id, cost)
    @id = id
    @cost = cost
  end
  def <=>(other)
    @cost <=> other.cost
  end
end

def main
  n = gets.to_i
  array = Array.new
  n.times do
    array << gets.split(" ").map(&:to_i)[1..-1]
  end
  queue = Priority_queue.new
  minimum = Array.new(n, 1 / 0.0)
  minimum[0] = 0
  nowcost = 0
  position = 0
  while nowcost < minimum.max
    array[position][0].times do |i|
      queue << Node.new(array[position][1 + 2 * i], array[position][2 + 2 * i] + nowcost)
    end
    position = queue.gets
    while minimum[position.id] < position.cost
      position = queue.gets
    end
    position, nowcost = position.id, position.cost
    minimum[position] = nowcost
  end
  minimum.each_with_index{|item, idx|puts "#{idx} #{item}"}
end

main