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):

# # import heapq
# # from copy import deepcopy
from collections import deque
# # from collections import Counter
# # from itertools import accumulate
# # from itertools import permutations
# import numpy as np
# # import math

n, m, k = map(int, input().split())
# a = list(map(int, input().split()))
# n = int(input())

friend =[[] for _ in range(n)]
block = [[] for _ in range(n)]
for i in range(m):
    a,b = map(int, input().split())
    a,b = a-1,b-1
    friend[a].append(b); friend[b].append(a)

for i in range(k):
    c,d = map(int, input().split())
    c,d = c-1,d-1
    block[c].append(d); block[d].append(c)

def suggest(s):
    count = 0
    seen = [False] * n
    que = deque([s])
    while que:
        start = que.popleft()
        seen[start] = True
        for i in friend[start]:
            if seen[i]: continue
            else:
                que.append(i)

    for i in range(n):
        if i != s and seen[i] and (i not in friend[s]) and (i not in block[s]):
            count += 1

    return count

for i in range(n):
    print(suggest(i))






# # import heapq
# # from copy import deepcopy
from collections import deque
# # from collections import Counter
# # from itertools import accumulate
# # from itertools import permutations
# import numpy as np
# # import math

n, m, k = map(int, input().split())
# a = list(map(int, input().split()))
# n = int(input())

friend =[[] for _ in range(n)]
block = [[] for _ in range(n)]
for i in range(m):
    a,b = map(int, input().split())
    a,b = a-1,b-1
    friend[a].append(b); friend[b].append(a)

for i in range(k):
    c,d = map(int, input().split())
    c,d = c-1,d-1
    block[c].append(d); block[d].append(c)

def suggest(s):
    count = 0
    seen = [False] * n
    que = deque([s])
    while que:
        start = que.pop()
        seen[start] = True
        for i in friend[start]:
            if seen[i]: continue
            else:
                que.append(i)

    for i in range(n):
        if i != s and seen[i] and (i not in friend[s]) and (i not in block[s]):
            count += 1

    return count

for i in range(n):
    print(suggest(i))






# import heapq
# from copy import deepcopy
# from collections import deque
# from collections import Counter
# from itertools import accumulate
# from itertools import permutations
# import numpy as np
# import math


# a = list(map(int, input().split()))
# n = int(input())

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

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

uf = UnionFind(n)
friend =[[] for _ in range(n)]
block = [[] for _ in range(n)]
for i in range(m):
    a,b = map(int, input().split())
    a,b = a-1,b-1
    uf.union(a,b)
    friend[a].append(b); friend[b].append(a)

for i in range(k):
    c,d = map(int, input().split())
    c,d = c-1,d-1
    if uf.same(c,d):
        block[c].append(d); block[d].append(c)

w = [0] * n
for i in range(n):
    w[i] = uf.size(i) - len(friend[i]) - len(block[i]) - 1

print(*w)


# import heapq
# from copy import deepcopy
# from collections import deque
# from collections import Counter
# from itertools import accumulate
# from itertools import permutations
# import numpy as np
# import math

n, m, k = map(int, input().split())
# a = list(map(int, input().split()))
# n = int(input())

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

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

uf = UnionFind(n)
friend =[[] for _ in range(n)]
block = [[] for _ in range(n)]
for i in range(m):
    a,b = map(int, input().split())
    a,b = a-1,b-1
    uf.union(a,b)
    friend[a].append(b); friend[b].append(a)

for i in range(k):
    c,d = map(int, input().split())
    c,d = c-1,d-1
    if uf.same(c,d):
        block[c].append(d); block[d].append(c)

w = [0] * n
for i in range(n):
    w[i] = uf.size(i) - len(friend[i]) - len(block[i]) - 1

print(*w)

Verdicts: Time Limit Exceeded

Time Limit Exceeded

Runtime Error

Accepted