Problem: An SNS has N users - User 1, User 2, \cdots, User N.  Between these N users, there are some relationships - M _friendships_ and K _blockships_.  For each i = 1, 2, \cdots, M, there is a bidirectional friendship between User A_i and User B_i.  For each i = 1, 2, \cdots, K, there is a bidirectional blockship between User C_i and User D_i.  We define User a to be a _friend candidate_ for User b when all of the following four conditions are satisfied:    * a \neq b.   * There is not a friendship between User a and User b.   * There is not a blockship between User a and User b.   * There exists a sequence c_0, c_1, c_2, \cdots, c_L consisting of integers between 1 and N (inclusive) such that c_0 = a, c_L = b, and there is a friendship between User c_i and c_{i+1} for each i = 0, 1, \cdots, L - 1.  For each user i = 1, 2, ... N, how many friend candidates does it have?

Submission attempts (in order):
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 union(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())


class UnionFindLabel(UnionFind):
    def __init__(self, labels):
        assert len(labels) == len(set(labels))

        self.n = len(labels)
        self.parents = [-1] * self.n
        self.d = {x: i for i, x in enumerate(labels)}
        self.d_inv = {i: x for i, x in enumerate(labels)}

    def find_label(self, x):
        return self.d_inv[super().find(self.d[x])]

    def union(self, x, y):
        super().union(self.d[x], self.d[y])

    def size(self, x):
        return super().size(self.d[x])

    def same(self, x, y):
        return super().same(self.d[x], self.d[y])

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

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

def MAP(): return map(int, input().split())
def ZIP(n): return zip(*(MAP() for _ in range(n)))

N, M, K = list(map(int, input().split()))
friend_dict = {i: set() for i in range(1, N + 1)}
block_dict = {i: set() for i in range(1, N + 1)}
for i in range(M):
    node1, node2 = list(map(int, input().split()))
    friend_dict[node1].add(node2)
    friend_dict[node2].add(node1)
for i in range(K):
    node1, node2 = list(map(int, input().split()))
    block_dict[node1].add(node2)
    block_dict[node2].add(node1)

uf = UnionFindLabel(range(1, N + 1))
for node, friends in friend_dict.items():
    for friend_node in friends:
        if not uf.same(node, friend_node):
            uf.union(node, friend_node)

for node in range(1, N + 1):
    cnt = 0
    for mem_node in uf.members(node):
        if mem_node != node and mem_node not in friend_dict[node] and mem_node not in block_dict[node]:
            cnt += 1
    print(cnt)


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 union(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())


class UnionFindLabel(UnionFind):
    def __init__(self, labels):
        assert len(labels) == len(set(labels))

        self.n = len(labels)
        self.parents = [-1] * self.n
        self.d = {x: i for i, x in enumerate(labels)}
        self.d_inv = {i: x for i, x in enumerate(labels)}

    def find_label(self, x):
        return self.d_inv[super().find(self.d[x])]

    def union(self, x, y):
        super().union(self.d[x], self.d[y])

    def size(self, x):
        return super().size(self.d[x])

    def same(self, x, y):
        return super().same(self.d[x], self.d[y])

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

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

def MAP(): return map(int, input().split())
def ZIP(n): return zip(*(MAP() for _ in range(n)))

N, M, K = list(map(int, input().split()))
ans = [-1] * N
uf = UnionFindLabel(range(1, N + 1))
for i in range(M):
    node1, node2 = list(map(int, input().split()))
    ans[node1 - 1] -= 1
    ans[node2 - 1] -= 1
    if not uf.same(node1, node2):
        uf.union(node1, node2)
for i in range(N):
    ans[i] += uf.size(i + 1)
for i in range(K):
    node1, node2 = list(map(int, input().split()))
    if uf.same(node1, node2):
        ans[node1 - 1] -= 1
        ans[node2 - 1] -= 1
print(*ans)


Verdicts: Time Limit Exceeded

Accepted