Buggy python submission (verdict: Runtime Error):
N, M, K = map(int, input().split())

groups = [None] * N
visited = set()

# 連結成分を求める
def dfs(group, i):
  if i in visited:
    return
  
  visited.add(i)
  group.add(i)
  groups[i] = group
  # 他のノード
  for c in graph[i]:
    if not c in visited:
      dfs(group, c)
  

graph = [set() for _ in range(N)]

friends_count = [0] * N
block_count = [0] * N

for _ in range(M):
  a, b = map(int, input().split())
  a, b = a - 1, b - 1
  friends_count[a] += 1
  friends_count[b] += 1
  graph[a].add(b)
  graph[b].add(a)
  
for i in range(N):
  dfs(set(), i)
      
for _ in range(K):
  c, d = map(int, input().split())
  c, d = c - 1, d - 1
  
  # 同じ連結成分に含まれないブロック関係は無視してok
  if d in groups[c]:
    block_count[c] += 1
    block_count[d] += 1

candidates = []

for i in range(N): 
  candidates.append(
    len(groups[i]) - friends_count[i] - block_count[i] - 1
  )
    
print(' '.join(str(x) for x in candidates))

Fixed submission:
import sys
sys.setrecursionlimit(2 ** 30)
N, M, K = map(int, input().split())

groups = [None] * N
visited = set()

# 連結成分を求める
def dfs(group, i):
  if i in visited:
    return
  
  visited.add(i)
  group.add(i)
  groups[i] = group
  # 他のノード
  for c in graph[i]:
    if not c in visited:
      dfs(group, c)
  

graph = [set() for _ in range(N)]

friends_count = [0] * N
block_count = [0] * N

for _ in range(M):
  a, b = map(int, input().split())
  a, b = a - 1, b - 1
  friends_count[a] += 1
  friends_count[b] += 1
  graph[a].add(b)
  graph[b].add(a)
  
for i in range(N):
  dfs(set(), i)
      
for _ in range(K):
  c, d = map(int, input().split())
  c, d = c - 1, d - 1
  
  # 同じ連結成分に含まれないブロック関係は無視してok
  if d in groups[c]:
    block_count[c] += 1
    block_count[d] += 1

candidates = []

for i in range(N): 
  candidates.append(
    len(groups[i]) - friends_count[i] - block_count[i] - 1
  )
    
print(' '.join(str(x) for x in candidates))

Buggy python submission (verdict: Runtime Error):
class UnionFind:
  def __init__(self, size):
    self.parents = [x for x in range(size)]
    self.sizes = [1] * size
  
  def unite(self, a, b):
    if self.find(a) == self.find(b):
      return
    
    if self.size_of(a) > self.size_of(b):
      self.sizes[self.find(a)] += self.size_of(b)
      self.parents[self.find(b)] = self.find(a)
    else:
      self.sizes[self.find(b)] += self.size_of(a)
      self.parents[self.find(a)] = self.find(b)
  
  def is_same(self, a, b):
    return self.find(a) == self.find(b)
  
  def size_of(self, a):
    return self.sizes[self.find(a)]
  
  def find(self, a):
    if a == self.parents[a]:
      return a
    else:
      self.parents[a] = self.find(self.parents[a])
      return self.parents[a]


N, M, K = map(int, input().split())

u = UnionFind(N)

# 自分自身・友達・ブロック関係のいずれかの場合はフラグ立つ
relations = [1 << i for i in range(N)]

for _ in range(M):
  a, b = map(int, input().split())
  a, b = a - 1, b - 1
  relations[a] |= 1 << b
  relations[b] |= 1 << a
  u.unite(a, b)
  
for _ in range(K):
  c, d = map(int, input().split())
  c, d = c - 1, d - 1
  
  # 同じ連結成分に含まれないブロック関係は無視してok
  if u.is_same(c, d):
    relations[c] |= 1 << d
    relations[d] |= 1 << c

candidates = []

for i in range(N):
  # candidates.append(u.size_of(i) - bin(relations[i]).count('1'))
  candidates.append(u.size_of(i)
    
print(' '.join(str(x) for x in candidates))

Fixed submission:
import sys
sys.setrecursionlimit(2 ** 30)
N, M, K = map(int, input().split())

groups = [None] * N
visited = set()

# 連結成分を求める
def dfs(group, i):
  if i in visited:
    return
  
  visited.add(i)
  group.add(i)
  groups[i] = group
  # 他のノード
  for c in graph[i]:
    if not c in visited:
      dfs(group, c)
  

graph = [set() for _ in range(N)]

friends_count = [0] * N
block_count = [0] * N

for _ in range(M):
  a, b = map(int, input().split())
  a, b = a - 1, b - 1
  friends_count[a] += 1
  friends_count[b] += 1
  graph[a].add(b)
  graph[b].add(a)
  
for i in range(N):
  dfs(set(), i)
      
for _ in range(K):
  c, d = map(int, input().split())
  c, d = c - 1, d - 1
  
  # 同じ連結成分に含まれないブロック関係は無視してok
  if d in groups[c]:
    block_count[c] += 1
    block_count[d] += 1

candidates = []

for i in range(N): 
  candidates.append(
    len(groups[i]) - friends_count[i] - block_count[i] - 1
  )
    
print(' '.join(str(x) for x in candidates))

Buggy python submission (verdict: Time Limit Exceeded):
class UnionFind:
  def __init__(self, size):
    self.parents = [x for x in range(size)]
    self.sizes = [1] * size
  
  def unite(self, a, b):
    if self.find(a) == self.find(b):
      return
    
    if self.size_of(a) > self.size_of(b):
      self.sizes[self.find(a)] += self.size_of(b)
      self.parents[self.find(b)] = self.find(a)
    else:
      self.sizes[self.find(b)] += self.size_of(a)
      self.parents[self.find(a)] = self.find(b)
  
  def is_same(self, a, b):
    return self.find(a) == self.find(b)
  
  def size_of(self, a):
    return self.sizes[self.find(a)]
  
  def find(self, a):
    if a == self.parents[a]:
      return a
    else:
      self.parents[a] = self.find(self.parents[a])
      return self.parents[a]


N, M, K = map(int, input().split())

u = UnionFind(N)

# 自分自身・友達・ブロック関係のいずれかの場合はフラグ立つ
relations = [1 << i for i in range(N)]

for _ in range(M):
  a, b = map(int, input().split())
  a, b = a - 1, b - 1
  relations[a] |= 1 << b
  relations[b] |= 1 << a
  u.unite(a, b)
  
for _ in range(K):
  c, d = map(int, input().split())
  c, d = c - 1, d - 1
  
  # 同じ連結成分に含まれないブロック関係は無視してok
  if u.is_same(c, d):
    relations[c] |= 1 << d
    relations[d] |= 1 << c

candidates = []

for i in range(N):
  # candidates.append(u.size_of(i) - bin(relations[i]).count('1'))
  candidates.append(u.size_of(i) - bin(relations[i]).count('1'))
    
print(' '.join(str(x) for x in candidates))

Fixed submission:
import sys
sys.setrecursionlimit(2 ** 30)
N, M, K = map(int, input().split())

groups = [None] * N
visited = set()

# 連結成分を求める
def dfs(group, i):
  if i in visited:
    return
  
  visited.add(i)
  group.add(i)
  groups[i] = group
  # 他のノード
  for c in graph[i]:
    if not c in visited:
      dfs(group, c)
  

graph = [set() for _ in range(N)]

friends_count = [0] * N
block_count = [0] * N

for _ in range(M):
  a, b = map(int, input().split())
  a, b = a - 1, b - 1
  friends_count[a] += 1
  friends_count[b] += 1
  graph[a].add(b)
  graph[b].add(a)
  
for i in range(N):
  dfs(set(), i)
      
for _ in range(K):
  c, d = map(int, input().split())
  c, d = c - 1, d - 1
  
  # 同じ連結成分に含まれないブロック関係は無視してok
  if d in groups[c]:
    block_count[c] += 1
    block_count[d] += 1

candidates = []

for i in range(N): 
  candidates.append(
    len(groups[i]) - friends_count[i] - block_count[i] - 1
  )
    
print(' '.join(str(x) for x in candidates))

Buggy python submission (verdict: Time Limit Exceeded):
class UnionFind:
  def __init__(self, size):
    self.parents = [x for x in range(size)]
    self.sizes = [1] * size
  
  def unite(self, a, b):
    if self.find(a) == self.find(b):
      return
    
    if self.size_of(a) > self.size_of(b):
      self.sizes[self.find(a)] += self.size_of(b)
      self.parents[self.find(b)] = self.find(a)
    else:
      self.sizes[self.find(b)] += self.size_of(a)
      self.parents[self.find(a)] = self.find(b)
  
  def is_same(self, a, b):
    return self.find(a) == self.find(b)
  
  def size_of(self, a):
    return self.sizes[self.find(a)]
  
  def find(self, a):
    if a == self.parents[a]:
      return a
    else:
      self.parents[a] = self.find(self.parents[a])
      return self.parents[a]


N, M, K = map(int, input().split())

u = UnionFind(N)

# 自分自身・友達・ブロック関係のいずれかの場合はフラグ立つ
relations = [1 << i for i in range(N)]

for _ in range(M):
  a, b = map(int, input().split())
  a, b = a - 1, b - 1
  relations[a] |= 1 << b
  relations[b] |= 1 << a
  u.unite(a, b)
  
for _ in range(K):
  c, d = map(int, input().split())
  c, d = c - 1, d - 1
  
  # 同じ連結成分に含まれないブロック関係は無視してok
  if u.is_same(c, d):
    relations[c] |= 1 << d
    relations[d] |= 1 << c

candidates = []

for i in range(N):
  candidates.append(u.size_of(i) - bin(relations[i]).count('1'))
    
print(' '.join(str(x) for x in candidates))

Fixed submission:
import sys
sys.setrecursionlimit(2 ** 30)
N, M, K = map(int, input().split())

groups = [None] * N
visited = set()

# 連結成分を求める
def dfs(group, i):
  if i in visited:
    return
  
  visited.add(i)
  group.add(i)
  groups[i] = group
  # 他のノード
  for c in graph[i]:
    if not c in visited:
      dfs(group, c)
  

graph = [set() for _ in range(N)]

friends_count = [0] * N
block_count = [0] * N

for _ in range(M):
  a, b = map(int, input().split())
  a, b = a - 1, b - 1
  friends_count[a] += 1
  friends_count[b] += 1
  graph[a].add(b)
  graph[b].add(a)
  
for i in range(N):
  dfs(set(), i)
      
for _ in range(K):
  c, d = map(int, input().split())
  c, d = c - 1, d - 1
  
  # 同じ連結成分に含まれないブロック関係は無視してok
  if d in groups[c]:
    block_count[c] += 1
    block_count[d] += 1

candidates = []

for i in range(N): 
  candidates.append(
    len(groups[i]) - friends_count[i] - block_count[i] - 1
  )
    
print(' '.join(str(x) for x in candidates))

Buggy python submission (verdict: Time Limit Exceeded):
class UnionFind:
  def __init__(self, size):
    self.parents = [x for x in range(size)]
    self.sizes = [1] * size
  
  def unite(self, a, b):
    if self.find(a) == self.find(b):
      return
    
    if self.size_of(a) > self.size_of(b):
      self.sizes[self.find(a)] += self.size_of(b)
      self.parents[self.find(b)] = self.find(a)
    else:
      self.sizes[self.find(b)] += self.size_of(a)
      self.parents[self.find(a)] = self.find(b)
  
  def is_same(self, a, b):
    return self.find(a) == self.find(b)
  
  def size_of(self, a):
    return self.sizes[self.find(a)]
  
  def find(self, a):
    if a == self.parents[a]:
      return a
    else:
      return self.find(self.parents[a])


N, M, K = map(int, input().split())

u = UnionFind(N)

# 自分自身・友達・ブロック関係のいずれかの場合はフラグ立つ
relations = [1 << i for i in range(N)]

for _ in range(M):
  a, b = map(int, input().split())
  a, b = a - 1, b - 1
  relations[a] |= 1 << b
  relations[b] |= 1 << a
  u.unite(a, b)
  
for _ in range(K):
  c, d = map(int, input().split())
  c, d = c - 1, d - 1
  
  # 同じ連結成分に含まれないブロック関係は無視してok
  if u.is_same(c, d):
    relations[c] |= 1 << d
    relations[d] |= 1 << c

candidates = []

for i in range(N):
  candidates.append(u.size_of(i) - bin(relations[i]).count('1'))
    
print(' '.join(str(x) for x in candidates))

Fixed submission:
import sys
sys.setrecursionlimit(2 ** 30)
N, M, K = map(int, input().split())

groups = [None] * N
visited = set()

# 連結成分を求める
def dfs(group, i):
  if i in visited:
    return
  
  visited.add(i)
  group.add(i)
  groups[i] = group
  # 他のノード
  for c in graph[i]:
    if not c in visited:
      dfs(group, c)
  

graph = [set() for _ in range(N)]

friends_count = [0] * N
block_count = [0] * N

for _ in range(M):
  a, b = map(int, input().split())
  a, b = a - 1, b - 1
  friends_count[a] += 1
  friends_count[b] += 1
  graph[a].add(b)
  graph[b].add(a)
  
for i in range(N):
  dfs(set(), i)
      
for _ in range(K):
  c, d = map(int, input().split())
  c, d = c - 1, d - 1
  
  # 同じ連結成分に含まれないブロック関係は無視してok
  if d in groups[c]:
    block_count[c] += 1
    block_count[d] += 1

candidates = []

for i in range(N): 
  candidates.append(
    len(groups[i]) - friends_count[i] - block_count[i] - 1
  )
    
print(' '.join(str(x) for x in candidates))

Buggy python submission (verdict: Time Limit Exceeded):
class UnionFind:
  def __init__(self, size):
    self.parents = [x for x in range(size)]
    self.sizes = [1] * size
  
  def unite(self, a, b):
    if self.find(a) == self.find(b):
      return
    
    if self.size_of(a) > self.size_of(b):
      self.sizes[self.find(a)] += self.size_of(b)
      self.parents[self.find(b)] = self.find(a)
    else:
      self.sizes[self.find(b)] += self.size_of(a)
      self.parents[self.find(a)] = self.find(b)
  
  def is_same(self, a, b):
    return self.find(a) == self.find(b)
  
  def size_of(self, a):
    return self.sizes[self.find(a)]
  
  def find(self, a):
    if a == self.parents[a]:
      return a
    else:
      return self.find(self.parents[a])


N, M, K = map(int, input().split())

u = UnionFind(N)

friendship = [[False] * (N) for _ in range(N)]
blocking = [[False] * (N) for _ in range(N)]

for _ in range(M):
  a, b = map(int, input().split())
  a, b = a - 1, b - 1
  friendship[a][b] = True
  friendship[b][a] = True
  u.unite(a, b)
  
for _ in range(K):
  c, d = map(int, input().split())
  c, d = c - 1, d - 1
  blocking[c][d] = True
  blocking[d][c] = True

candidates = []
# 10^10 :cry
for n in range(N):
  candidates.append(sum(
    m != n and not friendship[n][m] and not blocking[n][m] and u.is_same(n, m)
    for m in range(N)
  ))
    
print(' '.join(str(x) for x in candidates))

Fixed submission:
import sys
sys.setrecursionlimit(2 ** 30)
N, M, K = map(int, input().split())

groups = [None] * N
visited = set()

# 連結成分を求める
def dfs(group, i):
  if i in visited:
    return
  
  visited.add(i)
  group.add(i)
  groups[i] = group
  # 他のノード
  for c in graph[i]:
    if not c in visited:
      dfs(group, c)
  

graph = [set() for _ in range(N)]

friends_count = [0] * N
block_count = [0] * N

for _ in range(M):
  a, b = map(int, input().split())
  a, b = a - 1, b - 1
  friends_count[a] += 1
  friends_count[b] += 1
  graph[a].add(b)
  graph[b].add(a)
  
for i in range(N):
  dfs(set(), i)
      
for _ in range(K):
  c, d = map(int, input().split())
  c, d = c - 1, d - 1
  
  # 同じ連結成分に含まれないブロック関係は無視してok
  if d in groups[c]:
    block_count[c] += 1
    block_count[d] += 1

candidates = []

for i in range(N): 
  candidates.append(
    len(groups[i]) - friends_count[i] - block_count[i] - 1
  )
    
print(' '.join(str(x) for x in candidates))