bfs con colori

def opposite(i):
    if i == 0:
        return 1
    if i == 1:
        return 0

def BFS(G,s):
    n = size(G)
    reached = []
    color = []
    for i in range(n):
        reached.append(False)
        color.append(-1)
    tobeConsidered = [s]
    color[s] = 0
    reached[s] = True
    T = []
    i = 0
    while i < len(tobeConsidered):
        k = tobeConsidered[i]
        V = neighbors(G,k)
        for [j,w] in V:

            if color[j] == -1:
                color[j] = opposite(color[k])

            elif color[j] == color[k]:
                return False


            if reached[j] == False:
                T.append([k,j])
                reached[j] = True
                tobeConsidered.append(j)
        i = i+1

    return True