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):
#template
def inputlist(): return [int(j) for j in input().split()]
from collections import Counter
#template

#Union Find
N,M,K = inputlist()
graph = [[] for _ in range(N)]

def find(x):
    if par[x] < 0:
        return x
    else:
        par[x] = find(par[x])
        return par[x]

def unite(x,y):
    x = find(x)
    y = find(y)

    if x == y:
        return False
    else:
        
        if par[x] > par[y]:
            x,y = y,x
        par[x] += par[y]
        par[y] = x
        return True


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


def size(x):
    return -par[find(x)]



par = [-1]*(N)
for i in range(M):
    A,B = inputlist()
    graph[A-1].append(B-1)
    graph[B-1].append(A-1)
    unite(A-1,B-1)
bgraph = [[] for _ in range(N)]
for i in range(K):
    C,D = inputlist()
    if same(C-1,D-1):
        bgraph[C-1].append(D-1)
        bgraph[D-1].append(C-1)
li = [0]*N
for i in range(N):
    k = size(i)
    n = len(graph[i])
    m = len(bgraph[i])
    print(k,m,n)
    li[i] = k-n-m-1
print(*li)

#template
def inputlist(): return [int(j) for j in input().split()]
from collections import Counter
#template

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]

N,M,K = inputlist()
uf = UnionFind(N+1)
graph = [[] for _ in range(N+1)]
for i in range(M):
    a,b = inputlist()
    uf.union(a,b)
    graph[a].append(b)
    graph[b].append(a)
brogra = [[] for _ in range(N+1)]
li = [0]*N
for i in range(K):
    c,d = inputlist()
    if uf.same(c,d):
        brogra[c].append(d)
        brogra[d].append(c)
for i in range(1,N+1):
    li[i-1] = uf.size(i) - len(graph[i]) - len(brogra[i]) -1
print(*li)

Verdicts: Wrong Answer

Accepted