output_description: If the graph contains a negative cycle (a cycle whose sum of edge costs is a
  negative value) which is reachable from the source r, print
  
      
      
      NEGATIVE CYCLE
      
  
  in a line.
  
  Otherwise, print
  
      
      
      c0
      c1
      :
      c|V|-1
      
  
  The output consists of |V| lines. Print the cost of the shortest path from the
  source r to each vertex 0, 1, ... |V|-1 in order. If there is no path from the
  source to a vertex, print "INF".
input_description: An edge-weighted graph G (V, E) and the source r.
  
      
      
      |V| |E| r
      s0 t0 d0
      s1 t1 d1
      :
      s|E|-1 t|E|-1 d|E|-1
      
  
  |V| is the number of vertices and |E| is the number of edges in G. The graph
  vertices are named with the numbers 0, 1,..., |V|-1 respectively. r is the
  source of the graph.
  
  si and ti represent source and target vertices of i-th edge (directed) and di
  represents the cost of the i-th edge.
attempt: import sys
  
  fin = sys.stdin.readline
  
  
  MAX_NUM = float("inf")
  
  
  def initialize_int(adj, s):
      """
      assume that all vertex in {0, 1, 2, ..., N - 1}
      """
      N = len(adj)
      d = [MAX_NUM] * N
      d[s] = 0
      parent = [None] * N
      return d, parent
  
  
  def relax(d, w, source_v, target_v, parent):
      new_cost = d[source_v] + w[source_v][target_v]
      if new_cost < d[target_v]:
          d[target_v] = new_cost
          parent[target_v] = source_v
  
  
  # Even if a negative weight exists, bellman-ford can find shortest paths.
  # it also detects negative cycles,
  # but not every edges that consist the negative cycles is stored.
  # time complexity: O(VE + (V^2))
  def bellman_ford(adj, w, s):
      d, parent = initialize_int(adj, s)
  
      # calculate shortest paths
      for _ in range(len(adj)):
          for vertex in range(len(adj)):
              for neighbor in adj[vertex]:
                  relax(d, w, vertex, neighbor, parent)
  
      # detects negative cycles if exist
      negative_cycle_edges = set()
      for vertex in range(len(adj)):
          for neighbor in adj[vertex]:
              if d[neighbor] > d[vertex] + w[vertex][neighbor]:
                  negative_cycle_edges.add((vertex, neighbor))
      return d, parent, negative_cycle_edges
  
  
  V, E, r = [int(elem) for elem in fin().split()]
  adj = [[] for _ in range(V)]
  w = [[None] * V for _ in range(V)]
  for _ in range(E):
      s, t, d = [int(elem) for elem in fin().split()]
      adj[s].append(t)
      w[s][t] = d
  
  d, parent, negative_cycle_edges = bellman_ford(adj, w, r)
  if len(negative_cycle_edges) > 0:
      print("NEGATIVE CYCLE")
  else:
      for i in range(V):
          print("INF") if d[i] == MAX_NUM else print(d[i])
  
problem_description: Single Source Shortest Path (Negative Edges)
samples: [{"input": "4 5 0\n    0 1 2\n    0 2 3\n    1 2 -5\n    1 3 1\n    2 3 2", "output": "0\n    2\n    -3\n    -1"}, {"input": "4 6 0\n    0 1 2\n    0 2 3\n    1 2 -5\n    1 3 1\n    2 3 2\n    3 1 0", "output": "NEGATIVE CYCLE"}, {"input": "4 5 1\n    0 1 2\n    0 2 3\n    1 2 -5\n    1 3 1\n    2 3 2", "output": "INF\n    0\n    -5\n    -3"}]