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

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

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

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


N, M, K = map(int, input().split())
uf = UnionFind(N)
friends = [[] for _ in range(N)]
blocks = [[] for _ in range(N)]
root = {}

for i in range(M):
    a, b = map(int, input().split())
    friends[a - 1].append(b - 1)
    friends[b - 1].append(a - 1)
    uf.unite(a, b)

roots = uf.roots()
for r in roots:
    size = uf.size(r)
    root[r] = size - 1

for i in range(K):
    c, d = map(int, input().split())
    if uf.same(c - 1, d - 1):
        blocks[c - 1].append(d - 1)
        blocks[d - 1].append(c - 1)

ans = []
for i in range(N):
    ans.append(root[uf.find(i)] - len(blocks[i]) - len(friends[i]))
print(*ans)


N, M, K = map(int, input().split())
table = [['*'] * N for i in range(N)]
for _ in range(M + K):
  a, b = map(int, input().split())
  table[a - 1][b - 1] = ""
  table[b - 1][a - 1] = ""

for i in range(N):
  for j in range(N):
    if i == j:
      table[i][j] = ""
  print("".join(table[i]).count('*'), end=' ')



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

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

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

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


N, M, K = map(int, input().split())
uf = UnionFind(N)
friends = [[] for _ in range(N)]
blocks = [[] for _ in range(N)]
root = {}

for i in range(M):
    a, b = map(int, input().split())
    friends[a - 1].append(b - 1)
    friends[b - 1].append(a - 1)
    uf.unite(a - 1, b - 1)

for i in range(K):
    c, d = map(int, input().split())
    if uf.same(c - 1, d - 1):
        blocks[c - 1].append(d - 1)
        blocks[d - 1].append(c - 1)

ans = []
for i in range(N):
    ans.append(uf.size(i) - 1 - len(blocks[i]) - len(friends[i]))
print(*ans)


Verdicts: Runtime Error

Wrong Answer

Accepted