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 deque


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

ARR = []
BRR = []
for i in range(M):
    ARR.append(list(map(int,input().split())))

for i in range(K):
    BRR.append(list(map(int,input().split())))

#
# N = 10
# M = 9
# K = 3
# ARR = [
#     [10, 1],
#     [6, 7],
#     [8, 2],
#     [2, 5],
#     [8, 4],
#     [7, 3],
#     [10, 9],
#     [6, 4],
#     [5, 8],
#     [2, 6],
# ]
#
# BRR = [
#     [7, 5],
#     [3, 1],
# ]

def prepare(n, arr):
    nodes = [[] for i in range(n)]
    nodeStates = [False for i in range(n)]
    nodeLabels = [0 for i in range(n)]
    for ar in arr:
        fromNode = ar[0] - 1
        toNode = ar[1] - 1
        nodes[fromNode].append(toNode)
        nodes[toNode].append(fromNode)

    return nodes, nodeStates, nodeLabels

def prepare2(n,brr):
    nodes = [set() for i in range(n)]
    for ar in brr:
        fromNode = ar[0] - 1
        toNode = ar[1] - 1
        nodes[fromNode].add(toNode)

    return nodes


def calculate(n, m, k, nodes, nodeStates, nodeLabels,blockNodes):
    q = deque()

    label = 0
    labelDict = []
    while nodeStates.count(False) > 0:

        falseIndex = nodeStates.index(False)

        q.append(falseIndex)
        mmm = set()
        while len(q) > 0:
            node = q.popleft()
            if nodeStates[node] == True:
                continue
            nodeStates[node] = True
            nodeLabels[node] = label
            childNodes = nodes[node]
            mmm.add(node)
            for childNode in childNodes:
                if nodeStates[childNode] == False:
                    q.append(childNode)

        labelDict.append(mmm)
        label = label + 1

    result = []
    for i in range(n):
        l = nodeLabels[i]

        p = labelDict[l]
        pp = blockNodes[i]
        s = p & pp
        result.append(str(len(labelDict[l]) - len(nodes[i]) - 1 - len(s)))

    print(" ".join(result))


nodes, nodeStates, nodeLabels = prepare(N, ARR)
blockNodes = prepare2(N,BRR)
calculate(N, M, K, nodes, nodeStates, nodeLabels,blockNodes)


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 = 10
# M = 9
# K = 3
# ARR = [
#     [10, 1],
#     [6, 7],
#     [8, 2],
#     [2, 5],
#     [8, 4],
#     [7, 3],
#     [10, 9],
#     [6, 4],
#     [5, 8],
# ]
#
# BRR = [
#     [2, 6],
#     [7, 5],
#     [3, 1],
# ]
#



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

ARR = []

for i in range(M):
    ARR.append(list(map(int,input().split())))

BRR = []
for i in range(K):
    BRR.append(list(map(int,input().split())))

unionFind = UnionFind(N)

fl = [[] for i in range(N)]
bl = [[] for i in range(N)]
for ar in ARR:
    fromNode = ar[0] - 1
    toNode = ar[1] - 1
    unionFind.union(fromNode, toNode)
    fl[fromNode].append(toNode)
    fl[toNode].append(fromNode)

for br in BRR:
    fromNode = br[0] - 1
    toNode = br[1] - 1
    bl[fromNode].append(toNode)
    bl[toNode].append(fromNode)

result = []
for i in range(N):
    size = unionFind.size(i) - 1


    for f in fl[i]:
        if unionFind.same(i,f):
            size = size - 1
    for b in bl[i]:
        if unionFind.same(i,b):
            size = size - 1

    result.append(str(size))

print(" ".join(result))


Verdicts: Wrong Answer

Accepted