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 math
import itertools
import fractions
import heapq
import collections
import bisect
import sys
import copy

sys.setrecursionlimit(10**9)
mod = 10**7+9
inf = 10**20

def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(): return [list(map(int, l.split())) for l in sys.stdin.readlines()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return 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 size(self, x):
        return -self.parents[self.find(x)]

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


n,m,k = LI()

ab = [LI() for _ in range(m)]
cd = [LI() for _ in range(k)]
uf = UnionFind(n)

for a,b in ab:
    uf.union(a-1,b-1)
ans = [uf.size(i)-1 for i in range(n)]
print(uf.parents)
for a,b in ab:
    if uf.same(a-1,b-1):
        ans[a-1]-=1
        ans[b-1]-=1
for c,d in cd:
    if uf.same(c-1,d-1):
        ans[c-1]-=1
        ans[d-1]-=1

print(ans)

import math
import itertools
import fractions
import heapq
import collections
import bisect
import sys
import copy

sys.setrecursionlimit(10**9)
mod = 10**7+9
inf = 10**20

def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(): return [list(map(int, l.split())) for l in sys.stdin.readlines()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return 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 size(self, x):
        return -self.parents[self.find(x)]

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


n,m,k = LI()

ab = [LI() for _ in range(m)]
cd = [LI() for _ in range(k)]
uf = UnionFind(n)

for a,b in ab:
    uf.union(a-1,b-1)
ans = [uf.size(i)-1 for i in range(n)]
for a,b in ab:
    if uf.same(a-1,b-1):
        ans[a-1]-=1
        ans[b-1]-=1
for c,d in cd:
    if uf.same(c-1,d-1):
        ans[c-1]-=1
        ans[d-1]-=1

print(*ans)

Verdicts: Wrong Answer

Accepted