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=0):
        self.d = [-1 for _ in range(n)]
    
    def find(self, x): # find root
        if self.d[x] < 0: return x
        self.d[x] = self.find(self.d[x])
        return self.d[x]

    def unite(self, x, y):
        x, y = self.find(x), self.find(y)
        if x == y: return False
        if -self.d[x] < -self.d[y]: # swap
            t = x
            x = y
            y = t
        self.d[x] += self.d[y]
        self.d[y] = x
        return True

    def same(self, x, y):
        return self.find(x) == self.find(y)
    
    def size(self, x):
        return -self.d[self.find(x)]
    

def main():
    N, M, K = map(int, input().split())
    uf = UnionFind(N)

    deg = [0 for _ in range(N)]
    for _ in range(M): # friends
        a, b = map(int, input().split())
        a -= 1
        b -= 1
        deg[a] += 1
        deg[b] += 1
        uf.unite(a, b)
    print('----')
    print(uf.d)


    to = [[] for _ in range(N)]
    for _ in range(K): # block
        c, d = map(int, input().split())
        c -= 1
        d -= 1
        to[c].append(c)
        to[d].append(d)

    ret = []
    for i in range(N):
        ans = uf.size(i) - 1 - deg[i]
        for u in to[i]:
            if uf.same(i, u): ans -= 1
        ret.append(str(ans))

    print(' '.join(ret))


main()

from collections import defaultdict
class UnionFind:
    def __init__(self, n=0):
        self.d = [-1 for _ in range(n)]
    
    def find(self, x): # find root
        if self.d[x] < 0: return x
        self.d[x] = self.find(self.d[x])
        return self.d[x]

    def unite(self, x, y):
        x, y = self.find(x), self.find(y)
        if x == y: return False
        if -self.d[x] < -self.d[y]: # swap
            t = x
            x = y
            y = t
        self.d[x] += self.d[y]
        self.d[y] = x
        return True

    def same(self, x, y):
        return self.find(x) == self.find(y)
    
    def size(self, x):
        return -self.d[self.find(x)]
    

def main():
    N, M, K = map(int, input().split())
    uf = UnionFind(N)

    deg = [0 for _ in range(N)]
    for _ in range(M): # friends
        a, b = map(int, input().split())
        a -= 1
        b -= 1
        deg[a] += 1
        deg[b] += 1
        uf.unite(a, b)
    

    to = [[] for _ in range(N)]
    for _ in range(K): # block
        c, d = map(int, input().split())
        c -= 1
        d -= 1
        to[c].append(d)
        to[d].append(c)

    ret = []
    for i in range(N):
        ans = uf.size(i) - 1 - deg[i]
        for u in to[i]:
            if uf.same(i, u): ans -= 1
        ret.append(str(ans))

    print(' '.join(ret))


main()

Verdicts: Wrong Answer

Accepted