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()}

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

friend =[[] for _ in range(N)]
block = [[] for _ in range(N)]
for _ in range(M):
    a, b = map(int, input().split())
    a, b = a-1, b-1
    uf.union(a, b)
    friend[a].append(b)
    friend[b].append(a)

for i in range(k):
    c, d = map(int, input().split())
    c, d = c-1, d-1
    if uf.same(c, d):
        block[c].append(d)
        block[d].append(c)
ans = [0] * N
for i in range(N):
    ans[i] = uf.size(i) - len(friend[i]) - len(block[i]) - 1
print(*ans)


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()}

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

friend =[[] for _ in range(N)]
block = [[] for _ in range(N)]
for _ in range(M):
    a, b = map(int, input().split())
    a, b = a-1, b-1
    uf.union(a, b)
    friend[a].append(b)
    friend[b].append(a)

for _ in range(K):
    c, d = map(int, input().split())
    c, d = c-1, d-1
    if uf.same(c, d):
        block[c].append(d)
        block[d].append(c)
ans = [0] * N
for i in range(N):
    ans[i] = uf.size(i) - len(friend[i]) - len(block[i]) - 1
print(*ans)

Verdicts: Runtime Error

Accepted