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):

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 issame(self, x, y):
        return self.find(x) == self.find(y)

    def count(self,x):
        return -self.parents[self.find(x)]

    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 __str__(self):
        return '\n'.join('{}: {}'.format(root, self.members(root)) for root in self.roots())

n,m,k = map(int, input().split(" "))

uf_friend = UnionFind(n+1)
uf_block = UnionFind(n+1)
count = [0]*(n+1)
for i in range(m):
    a,b = map(int, input().split(" "))
    uf_friend.union(a, b)
    count[a] += 1
    count[b] += 1

for i in range(k):
    a,b = map(int, input().split(" "))
    if(uf_friend.issame(a,b)):
        count[a] += 1
        count[b] += 1

result = [0]*n
for i in range(1,n+1):
    result[i-1] = uf_friend.count(i) - count[i] - 1

print(result)


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 issame(self, x, y):
        return self.find(x) == self.find(y)

    def count(self,x):
        return -self.parents[self.find(x)]

    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 __str__(self):
        return '\n'.join('{}: {}'.format(root, self.members(root)) for root in self.roots())

n,m,k = map(int, input().split(" "))

uf_friend = UnionFind(n+1)
uf_block = UnionFind(n+1)
count = [0]*(n+1)
for i in range(m):
    a,b = map(int, input().split(" "))
    uf_friend.union(a, b)
    count[a] += 1
    count[b] += 1

for i in range(k):
    a,b = map(int, input().split(" "))
    if(uf_friend.issame(a,b)):
        count[a] += 1
        count[b] += 1

result = [0]*n
for i in range(1,n+1):
    result[i-1] = uf_friend.count(i) - count[i] - 1

print(*result)

Verdicts: Wrong Answer

Accepted