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):
＃@YottyPG　Qiita
from collections import deque

N,M,K = map(int,input().split())
friend = [list(map(int,input().split())) for _ in range(M)]
block = [list(map(int,input().split())) for _ in range(K)]

f = [set() for _ in range(N+1)]
b = [set() for _ in range(N+1)]

for i,j in friend:
    f[i].add(j)
    f[j].add(i)
for i,j in block:
    b[i].add(j)
    b[j].add(i)

stack=deque()
ans = [0]*(N+1)
visited = [0]*(N+1)

for i in range(1,N+1):
    if visited[i]:
        continue
    link = {i}
    visited[i] = 1 
    stack.append(i)
    while stack:
        n = stack.pop()
        for j in f[n]:
            if visited[j]==0:
                stack.append(j)
                visited[j] = 1
                link.add(j)
    for i in link:
        ans[i] = len(link) - len(link & f[i]) - len(link & b[i]) - 1
print(*ans[1:])


class UnionFind:
    def __init__(self, n):
        self.n = n
        self.parent = [i for i in range(n)]  
        self.rank = [1] * n  
        self.size = [1] * n  

    def find(self, x):  
        if self.parent[x] == x:
            return x
        else:
            self.parent[x] = self.find(self.parent[x])  
            return self.parent[x]

    def unite(self, x, y):  
        x = self.find(x)
        y = self.find(y)
        if x != y:
            if self.rank[x] < self.rank[y]:
                self.parent[x] = y
                self.size[y] += self.size[x]
            else:
                self.parent[y] = x
                self.size[x] += self.size[y]
                if self.rank[x] == self.rank[y]:
                    self.rank[x] += 1

    def is_same(self, x, y):  
        return self.find(x) == self.find(y)

    def group_size(self, x):  
        return self.size[self.find(x)]


N, M, K = map(int, input().split())
A, B = [0] * M, [0] * M  
C, D = [0] * K, [0] * K  

direct =[[] for _ in range(N)]  
uf = UnionFind(N)  

for i in range(M):
    A[i], B[i] = map(int, input().split())
    A[i] -= 1
    B[i] -= 1
    direct[A[i]].append(B[i])
    direct[B[i]].append(A[i])
    uf.unite(A[i], B[i])

for i in range(K):
    C[i], D[i] = map(int, input().split())
    C[i] -= 1
    D[i] -= 1
    if uf.is_same(C[i], D[i]):
        direct[C[i]].append(D[i])
        direct[D[i]].append(C[i])

ans = [0] * N
for i in range(N):
    ans[i] = uf.group_size(i) - len(direct[i]) - 1
print(*ans)

Verdicts: Runtime Error

Accepted