Problem:
On a planet far, far away, M languages are spoken. They are conveniently numbered 1 through M.  For _CODE FESTIVAL 20XX_ held on this planet, N participants gathered from all over the planet.  The i-th (1≦i≦N) participant can speak K_i languages numbered L_{i,1}, L_{i,2}, ..., L_{i,{}K_i}.  Two participants A and B can _communicate_ with each other if and only if one of the following conditions is satisfied:    * There exists a language that both A and B can speak.   * There exists a participant X that both A and B can communicate with.  Determine whether all N participants can communicate with all other participants.

Incorrect submission (verdict: Wrong Answer):
class UnionFind():
    def __init__(self, n):
        self.n = n
        self.parents = [-1] * n

    def find(self, x):
        if self.parents[x] < 0:
            return x
        else:
            self.parents[x] = self.find(self.parents[x])
            return self.parents[x]

    def unite(self, x, y):
        x = self.find(x)
        y = self.find(y)

        if x == y:
            return

        if self.parents[x] > self.parents[y]:
            x, y = y, x

        self.parents[x] += self.parents[y]
        self.parents[y] = x

    def size(self, x):
        return -self.parents[self.find(x)]

    def same(self, x, y):
        return self.find(x) == self.find(y)

    def members(self, x):
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root]

    def roots(self):
        return [i for i, x in enumerate(self.parents) if x < 0]

    def group_count(self):
        return len(self.roots())

    def all_group_members(self):
        return {r: self.members(r) for r in self.roots()}

    def __str__(self):
        return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())
n,m=map(int,input().split())
tree = UnionFind(n+m+1) 
for i in range(n):
  l=list(map(int,input().split()))
  for j in range(1,len(l)):
    tree.unite(i,l[j]+n)
rt=tree.find(0)
# print(tree)
for i in range(1,n):
  print(rt, tree.find(i))
  if rt!= tree.find(i):
    print("NO");exit()
print("YES")

Fixed, accepted submission:
class UnionFind():
    def __init__(self, n):
        self.n = n
        self.parents = [-1] * n

    def find(self, x):
        if self.parents[x] < 0:
            return x
        else:
            self.parents[x] = self.find(self.parents[x])
            return self.parents[x]

    def unite(self, x, y):
        x = self.find(x)
        y = self.find(y)

        if x == y:
            return

        if self.parents[x] > self.parents[y]:
            x, y = y, x

        self.parents[x] += self.parents[y]
        self.parents[y] = x

    def size(self, x):
        return -self.parents[self.find(x)]

    def same(self, x, y):
        return self.find(x) == self.find(y)

    def members(self, x):
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root]

    def roots(self):
        return [i for i, x in enumerate(self.parents) if x < 0]

    def group_count(self):
        return len(self.roots())

    def all_group_members(self):
        return {r: self.members(r) for r in self.roots()}

    def __str__(self):
        return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())
n,m=map(int,input().split())
tree = UnionFind(n+m+1) 
for i in range(n):
  l=list(map(int,input().split()))
  for j in range(1,len(l)):
    tree.unite(i,l[j]+n)
rt=tree.find(0)
# print(tree)
for i in range(1,n):
  # print(rt, tree.find(i))
  if rt!= tree.find(i):
    print("NO");exit()
print("YES")