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):
from collections import defaultdict

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 check():
    N,M,K = map(int, input().split())
    uf = UnionFind(N)
    friend = defaultdict(int)
    block = defaultdict(int)
    
    for _ in range(M):
        a,b = map(int, input().split())
        friend[a-1] += 1
        friend[b-1] += 1
        uf.union(a-1,b-1)
    
    for _ in range(K):
        a,b = map(int, input().split())
        if uf.same(a-1,b-1):
            block[a-1] += 1
            block[b-1] += 1
    
    result = [uf.size(i) - friend[i] - block[i] - 1 for i in range(N)]
    print(' '.join(map(str, result)))

from collections import defaultdict

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)
    


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

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

for _ in range(K):
    a,b = map(int, input().split())
    if uf.same(a-1,b-1):
        block[a-1] += 1
        block[b-1] += 1

result = [uf.size(i) - friend[i] - block[i] - 1 for i in range(N)]
print(' '.join(map(str, result)))

Verdicts: Wrong Answer

Accepted