Problem:


Score : 400 points

Problem Statement

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?

Constraints

All values in input are integers.

2 ≤ N ≤ 10^5

0 \leq M \leq 10^5

0 \leq K \leq 10^5

1 \leq A_i, B_i \leq N

A_i \neq B_i

1 \leq C_i, D_i \leq N

C_i \neq D_i

(A_i, B_i) \neq (A_j, B_j) (i \neq j)

(A_i, B_i) \neq (B_j, A_j)

(C_i, D_i) \neq (C_j, D_j) (i \neq j)

(C_i, D_i) \neq (D_j, C_j)

(A_i, B_i) \neq (C_j, D_j)

(A_i, B_i) \neq (D_j, C_j)

Input

Input is given from Standard Input in the following format:

N M K
A_1 B_1
\vdots
A_M B_M
C_1 D_1
\vdots
C_K D_K

Output

Print the answers in order, with space in between.

Sample Input 1

4 4 1
2 1
1 3
3 2
3 4
4 1

Sample Output 1

0 1 0 1

There is a friendship between User 2 and 3, and between 3 and 4. Also, there is no friendship or blockship between User 2 and 4. Thus, User 4 is a friend candidate for User 2.

However, neither User 1 or 3 is a friend candidate for User 2, so User 2 has one friend candidate.

Sample Input 2

5 10 0
1 2
1 3
1 4
1 5
3 2
2 4
2 5
4 3
5 3
4 5

Sample Output 2

0 0 0 0 0

Everyone is a friend of everyone else and has no friend candidate.

Sample Input 3

10 9 3
10 1
6 7
8 2
2 5
8 4
7 3
10 9
6 4
5 8
2 6
7 5
3 1

Sample Output 3

1 3 5 4 3 3 3 3 1 0



Sample input:
4 4 1
2 1
1 3
3 2
3 4
4 1


Sample output:
0 1 0 1

Submission status: Accepted

Solution code:
def main():
    import sys
    buf = sys.stdin.buffer

    class UnionFind:
        def __init__(self, size):
            self.parent = [-1] * size
            self.rank = [1] * size

        def get_root(self, node):
            parent = self.parent[node]
            if parent == -1:
                root = node
            else:
                root = self.get_root(parent)
                self.parent[node] = root  # 同じnodeへの2回目以降のget_rootを高速にするために、直接rootに繋いでおく
            return root

        def in_same_group(self, node1, node2):
            root1 = self.get_root(node1)
            root2 = self.get_root(node2)
            return root1 == root2

        def unite(self, node1, node2):
            if self.in_same_group(node1, node2):
                return
            main_root = self.get_root(node1)
            sub_root = self.get_root(node2)
            if self.rank[main_root] < self.rank[sub_root]:  # rankの大きい方をmain_rootにする
                main_root, sub_root = sub_root, main_root

            self.parent[sub_root] = main_root
            self.rank[main_root] += self.rank[sub_root]

    n, m, k = map(int, buf.readline().split())
    uf = UnionFind(n)
    minus_count = [1] * n

    query = list(map(int, buf.read().split()))
    follows = query[:2*m]
    blocks = query[2*m:]

    for a, b in zip(follows[0::2], follows[1::2]):
        uf.unite(a-1, b-1)
        minus_count[a-1] += 1
        minus_count[b-1] += 1

    for a, b in zip(blocks[0::2], blocks[1::2]):
        if uf.in_same_group(a-1, b-1):
            minus_count[a-1] += 1
            minus_count[b-1] += 1

    result = [uf.rank[uf.get_root(i)] - minus_count[i] for i in range(n)]
    print(*result)


if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
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)

    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 group_count(self):
        return len(self.roots())

    def all_group_members(self):
        return {r: self.members(r) for r in self.roots()}

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


N, M, K = map(int, input().split())
uf = UnionFind(N)
not_friend_candidates = [1]*N
for _ in range(M):
    A, B = map(int, input().split())
    uf.union(A-1, B-1)
    not_friend_candidates[A-1] += 1
    not_friend_candidates[B-1] += 1
for _ in range(K):
    C, D = map(int, input().split())
    if uf.same(C-1, D-1):
        not_friend_candidates[C-1] += 1
        not_friend_candidates[D-1] += 1
for hito in range(N):
    print(uf.size(hito) - not_friend_candidates[hito], end=" ")
print()

Submission status: Accepted

Solution code:
class UnionFind:
    def __init__(self, n):
        self.par =[i for i in range(n+1)]
        self.rank = [0] * (n+1)

    # 検索
    def find(self, x):
        if self.par[x] == x:
            return x
        else:
            self.par[x] = self.find(self.par[x])
            return self.par[x]

    # 併合
    def union(self, x, y):
        x = self.find(x)
        y = self.find(y)
        if self.rank[x] < self.rank[y]:
            self.par[x] = y
        else:
            self.par[y] = x
            if self.rank[x] == self.rank[y]:
                self.rank[x] += 1

    # 同じ集合に属するか判定
    def same_check(self, x, y):
        return self.find(x) == self.find(y)

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

friend=[0 for i in range(n+1)]
blocked={i:[] for i in range(1,n+1)}
for i in range(m):
    a,b=map(int,input().split())
    friend[a]+=1
    friend[b]+=1
    A.union(a,b)
    
    
for j in range(k):
    x,y=map(int,input().split())
    
    blocked[x].append(y)
    blocked[y].append(x)
    
for i in range(1,n+1):
    A.find(i)
    
parent=A.par

howmany={i:0 for i in range(n+1)}

for i in range(1,n+1):
    x=parent[i]
    howmany[parent[i]]+=1
    
#howmanyに同じグループが何個あるか

lists=[]
for i in range(1,n+1):
    sums=howmany[parent[i]]-1-friend[i]
    for some in blocked[i]:
        if parent[i]==parent[some]:
            sums-=1
    lists.append(sums)
print(*lists)

Submission status: Accepted

Solution code:
def main():
    import sys
    buf = sys.stdin.buffer

    class UnionFind:
        def __init__(self, size):
            self.parent = [-1] * size
            self.rank = [1] * size

        def get_root(self, node):
            parent = self.parent[node]
            if parent == -1:
                root = node
            else:
                root = self.get_root(parent)
                self.parent[node] = root  # 同じnodeへの2回目以降のget_rootを高速にするために、直接rootに繋いでおく
            return root

        def in_same_group(self, node1, node2):
            root1 = self.get_root(node1)
            root2 = self.get_root(node2)
            return root1 == root2

        def unite(self, node1, node2):
            if self.in_same_group(node1, node2):
                return
            main_root = self.get_root(node1)
            sub_root = self.get_root(node2)
            if self.rank[main_root] < self.rank[sub_root]:  # rankの大きい方をmain_rootにする
                main_root, sub_root = sub_root, main_root

            self.parent[sub_root] = main_root
            self.rank[main_root] += self.rank[sub_root]

    n, m, k = map(int, buf.readline().split())
    uf = UnionFind(n)
    minus_count = [1] * n

    def gen():
        for num in buf.read().split():
            yield int(num)

    generator = gen()
    for _ in range(m):
        a, b = generator.__next__(), generator.__next__()
        uf.unite(a-1, b-1)
        minus_count[a-1] += 1
        minus_count[b-1] += 1

    for i in range(k):
        a, b = generator.__next__(), generator.__next__()
        if uf.in_same_group(a-1, b-1):
            minus_count[a-1] += 1
            minus_count[b-1] += 1

    result = []
    for i in range(n):
        parent = uf.get_root(i)
        count = uf.rank[parent]
        ans = count - minus_count[i]
        result.append(ans)

    print(*result)

if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
class UnionFind():
    def __init__(self, n):
        self.n = n
        # 親の名(いない場合は自分)
        self.parent = list(range(n))
        # 子の数
        self.count = [0] * n

    def root(self, x):
        # xの根を返す
        if self.parent[x] == x:
            return x
        else:
            return self.root(self.parent[x])

    def union(self, x, y):
        # xとyをまとめる
        x0 = self.root(x)
        y0 = self.root(y)
        if x0 == y0:
            return
        if x0 > y0:
            x0, y0 = y0, x0
        self.parent[y0] = x0
        self.count[x0] += self.count[y0] + 1

    def size(self, x):
        # xの属する木の大きさ
        return self.count[self.root(x)] + 1

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


N, M, K = map(int, input().split())
tree = UnionFind(N)
# 同じグループでの友達の数とブロックの数
F, B = [0] * N, [0] * N
lst = []
for m in range(M):
    a, b = map(int, input().split())
    tree.union(a - 1, b - 1)
    lst.append((a, b))
for a, b in lst:
    if tree.same(a - 1, b - 1):
        F[a - 1] += 1
        F[b - 1] += 1
for k in range(K):
    c, d = map(int, input().split())
    if tree.same(c - 1, d - 1):
        B[c - 1] += 1
        B[d - 1] += 1
# nが属する木の大きさから友達の数とブロックの数と自分を除いたのが答え
print(*[tree.size(n) - F[n] - B[n] - 1 for n in range(N)])

Submission status: Accepted

Solution code:
def main():
    import sys
    from operator import itemgetter
    buf = sys.stdin.buffer

    class UnionFind:
        def __init__(self, size):
            self.parent = [-1] * size
            self.rank = [1] * size

        def get_root(self, node):
            parent = self.parent[node]
            if parent == -1:
                root = node
            else:
                root = self.get_root(parent)
                self.parent[node] = root  # 同じnodeへの2回目以降のget_rootを高速にするために、直接rootに繋いでおく
            return root

        def in_same_group(self, node1, node2):
            root1 = self.get_root(node1)
            root2 = self.get_root(node2)
            return root1 == root2

        def unite(self, node1, node2):
            if self.in_same_group(node1, node2):
                return
            main_root = self.get_root(node1)
            sub_root = self.get_root(node2)
            if self.rank[main_root] < self.rank[sub_root]:  # rankの大きい方をmain_rootにする
                main_root, sub_root = sub_root, main_root

            self.parent[sub_root] = main_root
            self.rank[main_root] += self.rank[sub_root]

    n, m, k = map(int, buf.readline().split())
    uf = UnionFind(n)
    minus_count = [1] * n

    query = map(int, buf.read().split())

    def gen():
        for num in query:
            yield num

    generator = gen()
    for _ in range(m):
        a, b = generator.__next__(), generator.__next__()
        uf.unite(a-1, b-1)
        minus_count[a-1] += 1
        minus_count[b-1] += 1

    for i in range(k):
        a, b = generator.__next__(), generator.__next__()
        if uf.in_same_group(a-1, b-1):
            minus_count[a-1] += 1
            minus_count[b-1] += 1

    result = []
    for i in range(n):
        parent = uf.get_root(i)
        count = uf.rank[parent]
        ans = count - minus_count[i]
        result.append(ans)

    print(*result)

if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
from collections import deque
from copy import deepcopy

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)

    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 group_count(self):
        return len(self.roots())

    def all_group_members(self):
        return {r: self.members(r) for r in self.roots()}

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

N, M, K = map(int, input().split())
UF = UnionFind(N+1)
direct_connect = [0]*(N+1)
for _ in range(M):
  a, b = map(int, input().split())
  UF.union(a, b)
  direct_connect[a] += 1
  direct_connect[b] += 1
for _ in range(K):
  a, b = map(int, input().split())
  if UF.same(a, b):
    direct_connect[a] += 1
    direct_connect[b] += 1
print(" ".join([str(UF.size(i)-direct_connect[i]-1) for i in range(1, N+1)]))

Submission status: Accepted

Solution code:
import sys
from io import StringIO
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
import unittest

class UnionFind():
    # 作りたい要素数nで初期化
    # 使用するインスタンス変数の初期化
    def __init__(self, n):
        self.n = n
        # root[x]<0ならそのノードが根かつその値が木の要素数
        # rootノードでその木の要素数を記録する
        self.root = [-1]*(n+1)
        # 木をくっつける時にアンバランスにならないように調整する
        self.rnk = [0]*(n+1)

    # ノードxのrootノードを見つける
    def Find_Root(self, x):
        if(self.root[x] < 0):
            return x
        else:
            # ここで代入しておくことで、後の繰り返しを避ける
            self.root[x] = self.Find_Root(self.root[x])
            return self.root[x]
    # 木の併合、入力は併合したい各ノード
    def Unite(self, x, y):
        # 入力ノードのrootノードを見つける
        x = self.Find_Root(x)
        y = self.Find_Root(y)
        # すでに同じ木に属していた場合
        if(x == y):
            return 
        # 違う木に属していた場合rnkを見てくっつける方を決める
        elif(self.rnk[x] > self.rnk[y]):
            self.root[x] += self.root[y]
            self.root[y] = x

        else:
            self.root[y] += self.root[x]
            self.root[x] = y
            # rnkが同じ（深さに差がない場合）は1増やす
            if(self.rnk[x] == self.rnk[y]):
                self.rnk[y] += 1
    # xとyが同じグループに属するか判断
    def isSameGroup(self, x, y):
        return self.Find_Root(x) == self.Find_Root(y)

    # ノードxが属する木のサイズを返す
    def Count(self, x):
        return -self.root[self.Find_Root(x)]

def resolve():
    n,m,k=map(int, readline().split())
    #list(map(int,input().strip().split()))

    uf = UnionFind(n)

    res=[[] for i in range(n)]
    frd=[0 for i in range(n)]
    blk=[0 for i in range(n)]

    for i in range(m):
        #a,b=list(map(int,input().strip().split()))
        a,b=map(int, readline().split())
        a-=1
        b-=1
        uf.Unite(a,b)
        frd[a]+=1
        frd[b]+=1
    
    for i in range(k):
        #a,b=list(map(int,input().strip().split()))
        a,b=map(int, readline().split())
        a-=1
        b-=1
        if uf.isSameGroup(a,b):
            blk[a]+=1
            blk[b]+=1
    
    rescnt=[]

    for i in range(n):
        ri=uf.Count(i)-1-blk[i]-frd[i]
        rescnt.append(str(ri))
        ##rescnt.append(str(n-uf.size(i)-len(blk[i])))
    
    print(" ".join(rescnt))

resolve()

Submission status: Accepted

Solution code:
N, M, K = map(int, input().split())
AB = [list(map(int, input().split())) for _ in range(M)]
CD = [list(map(int, input().split())) for _ in range(K)]

root = [i for i in range(N + 1)]
size = [1] * (N + 1)
 
def find(x):
  y = root[x]
  if x == y:
    return x
  else:
    z = find(y)
    root[x] = z
    return z

def union(x, y):
  rx = find(x)
  ry = find(y)
  sx = size[rx]
  sy = size[ry]
  if rx == ry:
    return 0
  else:
    if sx >= sy:
      root[ry] = rx
      size[rx] = sx + sy
    else:
      root[rx] = ry
      size[ry] = sx + sy
    return sx * sy
    
for a, b in AB:
    union(a, b)

answer = []
for i in range(N + 1):
    answer.append(size[find(i)] - 1)

for a, b in AB:
    if find(a) == find(b):
        answer[a] -= 1
        answer[b] -= 1
for c, d in CD:
    if find(c) == find(d):
        answer[c] -= 1
        answer[d] -= 1
        
print(*answer[1:])

Submission status: Accepted

Solution code:
import sys
input = sys.stdin.buffer.readline

# https://note.nkmk.me/python-union-find/
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)

    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 group_count(self):
        return len(self.roots())

    def all_group_members(self):
        return {r: self.members(r) for r in self.roots()}

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


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

n_friend = [0]*N
for i in range(M):
    a, b = map(int, input().split())
    uf.union(a-1, b-1)
    n_friend[a-1] += 1
    n_friend[b-1] += 1

n_block_in_friend_cluster = [0]*N
for i in range(K):
    c, d = map(int, input().split())
    if uf.same(c-1, d-1):
        n_block_in_friend_cluster[c-1] += 1
        n_block_in_friend_cluster[d-1] += 1

ans = [uf.size(i) - n_friend[i] - n_block_in_friend_cluster[i] - 1 for i in range(N)]
print(*ans)

Submission status: Accepted

Solution code:

class UnionFindTree():
    # Union find tree
    # 2つの効率化　経路圧縮、ランク
    # つなげることはできるか分離することはできないので、木をつなげる方向に利用する
    # 
    def __init__(self, node_size):
        # node_num: 全体のノード数
        # parent: ノードがrootなら-1、親がいれば親の番号を入れる
        # size:
        # rank:  
        # self.parent =[-1] * node_size
        self.parent =[-1]*node_size
 
        self.size = [1] * node_size
        self.rank = [1] * node_size
    
    def find_root(self,node):
        # node属する木のrootを探す
        # たどった経路に含まれるnodeの親をrootにつなぎ替える
        if self.parent[node] == -1:             # 自身がrootの場合
            root = node
        else:                                   # 自身がrootでない場合
            root = self.find_root(self.parent[node]) # 再帰的にrootを探す
            self.parent[node] = root            # たどった経路に含まれるnodeの親をrootにつなぎ替える(経路圧縮)
        return root

    def unite(self,node0, node1):
        # node0とnode1が含まれる木を結合する
        
        #node0 node1のrootを求める
        root_node0 = self.find_root(node0) 
        root_node1 = self.find_root(node1)

        if root_node0 != root_node1:
            # rankを比較しrankが小さい方の木を大きい方の木のrootにつなげる
            # rank is sameのときのみrankが増える
            if self.rank[root_node0] == self.rank[root_node1]: # rank is same
                self.parent[root_node1] = root_node0
                self.rank[root_node0] += 1
                self.size[root_node0] += self.size[root_node1]

            elif self.rank[root_node0] > self.rank[root_node1]:
                self.parent[root_node1] = root_node0
                self.size[root_node0] += self.size[root_node1]

            else:
                self.parent[root_node0] = root_node1

                self.size[root_node1] += self.size[root_node0]

    def same(self,node0,node1):
        #同じグループか確かめる
        return self.find_root(node0)==self.find_root(node1)


    # def recombind(self,node):
    #     # rootがnodeのものを整理し直す


N,M,K = map(int,input().split())

ut = UnionFindTree(N)
friend_dict = {i:0 for i in range(N)}

for i in range(M):
    x,y = map(int,input().split())
    x-=1
    y-=1
    ut.unite(x,y)
    friend_dict[x]+=1
    friend_dict[y]+=1
    # friend_dict[x].extend([y])
    # friend_dict[y].extend([x])

# for i in range(N):
#     ut.find_root(i)

# block_dict = {i:[] for i in range(N)}
block_dict = {i:0 for i in range(N)}

for i in range(K):
    x,y = map(int,input().split())
    x-=1
    y-=1
    if ut.same(x,y):
        block_dict[x]+=1
        block_dict[y]+=1
        # block_dict[x].extend([y])
        # block_dict[y].extend([x])

ans =""
for i in range(N):
    group =  ut.find_root(i)
    group_size = ut.size[group]

    # cand = group_size - len(friend_dict[i]) - len(block_dict[i])
    cand = group_size - friend_dict[i] - block_dict[i] - 1
    ans += str(cand) + " "
print(ans)

Submission status: Accepted

Solution code:
def find(x):
    if par[x] < 0:
        return x
    else:
        par[x] = find(par[x])
        return par[x]

def same(x, y):
    if find(x) == find(y):
        return True
    else:
        return False

def unite(x, y):
    x = find(x)
    y = find(y)
    if x == y:
        return
    if par[x] > par[y]:
        x, y = y, x
    par[x] += par[y]
    par[y] = x

def uCnt(x):
    return -par[find(x)]

N, M, K = map(int, input().split())
par = [-1 for i in range(N + 1)]
rank = [0] * (N + 1)
frendCnt = [0] * (N + 1)
bloc = [[] for _ in range(N + 1)]
for i in range(M):
    a, b = map(int, input().split())
    a -= 1
    b -= 1
    frendCnt[a] += 1
    frendCnt[b] += 1
    unite(a, b)

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

for i in range(N):
    ans = uCnt(i) - frendCnt[i] - 1
    for j in bloc[i]:
      if same(i, j):
           ans -= 1
    print(ans, end=' ')

Submission status: Accepted

Solution code:
import sys
sys.setrecursionlimit(100000)

N,M, K = map(int,input().split())
AB = [list(map(int,input().split())) for i in range(M)]
CD = [list(map(int,input().split())) for i in range(K)]


#xの木の根を求める
def root(x,li): 
    if li[x] == x:
        return x
    else:
        li[x] = root(li[x],li)
        return li[x]

#xとyが同じ集合に属するかの判定
def find(x,y,li):
    return root(x,li) == root(y,li)

#xとｙを併合
def union(x,y,li):
    x = root(x,li)
    y = root(y,li)
    if x==y:
        return
    else:
        li[x]=y

li1 = [i for i in range(N)]
li2 = [i for i in range(N)]

for i in range(M):
    a = AB[i][0]
    b = AB[i][1]
    # print(str(i)+" ",end="")
    union(min(a-1,b-1),max(a-1,b-1),li1)
# for i in range(K):
#     c = CD[i][0]
#     d = CD[i][1]
#     union(c-1,d-1,li2)

mm = [0 for i in range(max(li1)+1)]
for i in range(N):
    m = root(i,li1)
    mm[m] += 1

# print(li1)
# print(mm)
ans = [mm[li1[i]]-1 for i in range(N)]
# print(ans)

for i in range(M):
    a = AB[i][0]
    b = AB[i][1]
    ans[a-1] -= 1
    ans[b-1] -= 1


for i in range(K):
    c = CD[i][0]
    d = CD[i][1]
    if find(c-1,d-1,li1):
        ans[c-1] -= 1
        ans[d-1] -= 1
print(*ans)

Submission status: Accepted

Solution code:
N,M,K=map(int,input().split())  
par=[i for i in range(N)]
rank=[0]*(N)
friend=[0]*N
block=[0]*N
size=[1]*N
def find(x):
  if par[x]==x:
    return x
  else:
    par[x]=find(par[x])
    return par[x]

#同じ集合か判定
def same(x,y):
  return find(x)==find(y)

def union(x,y):
  x=find(x)
  y=find(y)
  if x==y:
    return 
  if rank[x]>rank[y]:
    par[y]=x
    size[x]+=size[y]
  else:
    par[x]=y
    size[y]+=size[x]
    if rank[x]==rank[y]:
      rank[y]+=1


for i in range(M):
  a,b=map(int,input().split())
  a-=1
  b-=1
  union(a,b)
  friend[a]+=1
  friend[b]+=1

for i in range(K):
  c,d=map(int,input().split())
  c-=1
  d-=1
  if same(c,d):
    block[c]+=1
    block[d]+=1
    
ans=[]
for i in range(N):
  ans.append(size[find(i)]-friend[i]-block[i]-1)
print(*ans)

Submission status: Accepted

Solution code:
import sys
input = sys.stdin.readline

#Union Find

#xの根を求める
def find(x):
    if par[x] < 0:
        return x
    else:
        par[x] = find(par[x])
        return par[x]
#xとyの属する集合の併合
def unite(x,y):
    x = find(x)
    y = find(y)

    if x == y:
        return False
    else:
        #sizeの大きいほうがx
        if par[x] > par[y]:
            x,y = y,x
        par[x] += par[y]
        par[y] = x
        return True

#xとyが同じ集合に属するかの判定
def same(x,y):
    return find(x) == find(y)

#xが属する集合の個数
def size(x):
    return -par[find(x)]

N, M, K = map(int, input().split())
par = [-1 for i in range(N)]
rank = [0] * N
frendCnt = [0] * N
bloc = [[] for _ in range(N)]
for i in range(M):
    a, b = map(int, input().split())
    a -= 1
    b -= 1
    frendCnt[a] += 1
    frendCnt[b] += 1
    unite(a, b)

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

for i in range(N):
    ans = size(i) - frendCnt[i] - 1
    for j in bloc[i]:
      if same(i, j):
           ans -= 1
    print(ans, end=' ')

Submission status: Accepted

Solution code:
import sys
from io import StringIO
import unittest

class UnionFind():
    # 作りたい要素数nで初期化
    # 使用するインスタンス変数の初期化
    def __init__(self, n):
        self.n = n
        # root[x]<0ならそのノードが根かつその値が木の要素数
        # rootノードでその木の要素数を記録する
        self.root = [-1]*(n+1)
        # 木をくっつける時にアンバランスにならないように調整する
        self.rnk = [0]*(n+1)

    # ノードxのrootノードを見つける
    def Find_Root(self, x):
        if(self.root[x] < 0):
            return x
        else:
            # ここで代入しておくことで、後の繰り返しを避ける
            self.root[x] = self.Find_Root(self.root[x])
            return self.root[x]
    # 木の併合、入力は併合したい各ノード
    def Unite(self, x, y):
        # 入力ノードのrootノードを見つける
        x = self.Find_Root(x)
        y = self.Find_Root(y)
        # すでに同じ木に属していた場合
        if(x == y):
            return 
        # 違う木に属していた場合rnkを見てくっつける方を決める
        elif(self.rnk[x] > self.rnk[y]):
            self.root[x] += self.root[y]
            self.root[y] = x

        else:
            self.root[y] += self.root[x]
            self.root[x] = y
            # rnkが同じ（深さに差がない場合）は1増やす
            if(self.rnk[x] == self.rnk[y]):
                self.rnk[y] += 1
    # xとyが同じグループに属するか判断
    def isSameGroup(self, x, y):
        return self.Find_Root(x) == self.Find_Root(y)

    # ノードxが属する木のサイズを返す
    def Count(self, x):
        return -self.root[self.Find_Root(x)]

def resolve():
    n,m,k=list(map(int,input().strip().split()))

    uf = UnionFind(n)

    res=[[] for i in range(n)]
    frd=[0 for i in range(n)]
    blk=[0 for i in range(n)]

    for i in range(m):
        a,b=list(map(int,input().strip().split()))
        a-=1
        b-=1
        uf.Unite(a,b)
        frd[a]+=1
        frd[b]+=1
    
    for i in range(k):
        a,b=list(map(int,input().strip().split()))
        a-=1
        b-=1
        if uf.isSameGroup(a,b):
            blk[a]+=1
            blk[b]+=1
    
    rescnt=[]

    for i in range(n):
        ri=uf.Count(i)-1-blk[i]-frd[i]
        rescnt.append(str(ri))
        ##rescnt.append(str(n-uf.size(i)-len(blk[i])))
    
    print(" ".join(rescnt))

resolve()

Submission status: Accepted

Solution code:
import sys
int1 = lambda x: int(x) - 1
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
sys.setrecursionlimit(500000)


class UnionFind(object):

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

    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 group_count(self):
        return len(self.roots())

    def all_group_members(self):
        return {r: self.members(r) for r in self.roots()}


N, M, K = map(int, readline().split())

uf = UnionFind(N)
f = [0] * N
bl = [0] * N

for _ in range(M):
    a, b = map(int1, readline().split())
    f[a] += 1
    f[b] += 1
    uf.unite(a, b)

for i in range(K):
    c, d = map(int1, readline().split())
    if uf.find(c) == uf.find(d):
        bl[c] += 1
        bl[d] += 1

ans = [0] * N

for i in range(N):
    ans[i] = uf.size(i) - f[i] - bl[i] - 1

print(*ans)

Submission status: Accepted

Solution code:
#!/usr/bin/python3
# -*- coding:utf-8 -*-

from collections import defaultdict
import sys
sys.setrecursionlimit(10**5+10)

def main():
  n, m, k = map(int, input().strip().split())
  friends = defaultdict(lambda :{'children':list(), 'x':-1, 'visited':False})
  blocks = defaultdict(list)
  visited = set()
  
  for _ in range(m):
    a, b = map(int, input().strip().split())
    friends[a]['children'].append(b)
    friends[b]['children'].append(a)

  for _ in range(k):
    c, d = map(int, input().strip().split())
    blocks[c].append((d))
    blocks[d].append((c))
    
  def dfs(i, x):
    friends[i]['x'] = x
    friends[i]['visited'] = True
    count = 1
    for child in friends[i]['children']:
      if friends[child]['visited']:
        continue
      count += dfs(child, x)
    return count
  
  num = 1
  size = dict()
  for i in range(1, n+1):
    if friends[i]['visited']:
      continue
    count = dfs(i, num)
    size[num] = count
    num += 1
    
  ans = []
  for i in range(1, n+1):
    nblock = 0
    for ib in blocks[i]:
      if friends[i]['x'] == friends[ib]['x']:
        nblock += 1
    ans.append(size[friends[i]['x']] - len(friends[i]['children']) - nblock - 1)
  print(' '.join(list(map(str, ans))))
  
if __name__=='__main__':
  main()