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())


N, M, K = map(int, input().split())
uf = UnionFind(N)
not_friend_candidates = [1]*N
for _ in range(M):
    A, B = map(int, input().split())
    uf.union(A-1, B-1)
    not_friend_candidates[A-1] += 1
    not_friend_candidates[B-1] += 1
for _ in range(K):
    C, D = map(int, input().split())
    not_friend_candidates[C-1] += 1
    not_friend_candidates[D-1] += 1
for hito in range(N):
    print(uf.size(hito) - not_friend_candidates[hito], end=" ")
print()

import numpy as np
from scipy.sparse.csgraph import shortest_path
from scipy.sparse import csr_matrix
inf = 10**6


N, M, K = map(int, input().split())
graph = np.zeros((N, N), np.int32)
for _ in range(M):
    A, B = map(int, input().split())
    graph[A-1][B-1] = 1
    graph[B-1][A-1] = 1
for _ in range(K):
    C, D = map(int, input().split())
    graph[C-1][D-1] = inf
    graph[D-1][C-1] = inf


csgraph = csr_matrix(graph)
shortest_paths = shortest_path(csgraph, directed=False, method="BF")


ok = sorted(list(zip(*np.where(graph < inf))))
now = 0
ans = 0
for x, y in ok:
    if x == now:
        if shortest_paths[x][y] > 1:
            ans += 1
    else:
        print(ans, end=" ")
        ans = 0
        now += 1
        while now < x:
            if now != N - 1:
                print(0, end=" ")
            else:
                print(0, end="")
            now += 1
        if shortest_paths[x][y] != 1:
            ans += 1

while now <= N-1:
    if ans != 0:
        print(ans , end=" ")
        continue
    if now != N-1:
        print(0, end=" ")
    else:
        print(0, end="")
    now += 1
print()



import sys
input = sys.stdin.readline


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

        self.group_count -= 1

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

def int_(num_str):
    return int(num_str) - 1


N, M, K = map(int, input().split())
uf = UnionFind(N)
some = [1] * N
for i in range(M):
    a, b = map(int_, input().split())
    uf.union(a, b)
    some[a] += 1
    some[b] += 1
for i in range(K):
    c, d = map(int_, input().split())
    if uf.same(c, d):
        some[c] += 1
        some[d] += 1


ans = [uf.size(i) - some[i] for i in range(N)]
print(*ans)


Verdicts: Wrong Answer

Runtime Error

Accepted