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

Solution (Python):
import time, sys
sys.setrecursionlimit(int(1e7))

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

friends = [[] for i in range(n)]
cands = [None for i in range(n)]
cycles = []
visited = [False for i in range(n)]
count = [0 for i in range(n)]
for _ in range(m):
    a, b = map(lambda x: int(x)-1, input().split())
    friends[a].append(b)
    friends[b].append(a)

def dfs(x):
    visited[x] = True
    cycle.add(x)
    for i in friends[x]:
        if not visited[i]:
            dfs(i)
    return


for i, v in enumerate(visited):
    if v:
        continue
    cycle = set()
    dfs(i)
    l = len(cycle)
    ind = len(cycles)
    cycles.append(cycle)
    for fr in cycle:
        cands[fr] = ind
        count[fr] = l - 1 - len(friends[fr])


for _ in range(k):
    a, b = map(lambda x: int(x)-1, input().split())
    if a in cycles[cands[b]]:
        count[a] -= 1
        count[b] -= 1

for i in count:
    print(i, end=" ")

Solution (Python):
import sys
from collections import defaultdict
sys.setrecursionlimit(10 ** 6)

n,m,k=map(int, input().split())  
friend=[[] for _ in range(n)]

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

par=[None]*n
frmember=defaultdict(int)

def dfs(v):
    if par[i]==None:
        par[v]=v
        frmember[par[v]]+=1
    for f in friend[v]:
        if par[f]==None:
            par[f]=par[v]
            frmember[par[v]]+=1
            dfs(f)

for i in range(n):
    if par[i]==None:
        dfs(i)

ans=[0]*n
for i in range(n):
    ans[i]+=frmember[par[i]]-len(friend[i])-1

for i in range(k):
    a,b=map(int, input().split()) 
    if par[a-1]==par[b-1]:
        ans[a-1]-=1
        ans[b-1]-=1

print(*ans)

Solution (Python):
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)]


if __name__ == "__main__":
    n, m, k = map(int, input().split())

    u = UnionFind(n)
    f_cnt = [0] * n
    b_cnt = [0] * n

    for _ in range(m):
        a, b = map(int, input().split())
        u.union(a - 1, b - 1)
        f_cnt[a - 1] += 1
        f_cnt[b - 1] += 1

    for _ in range(k):
        c, d = map(int, input().split())
        if u.same(c - 1, d - 1):
            b_cnt[c - 1] += 1
            b_cnt[d - 1] += 1

    print(*[(u.size(i) - f_cnt[i] - b_cnt[i] - 1) for i in range(n)])

Solution (Python):
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 = map(int,input().split())
Friend_Net = UnionFind(N)
Block = [[] for i in range(N)]
friend = [0] * N

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

for i in range(K):
    C, D = map(int, input().split())
    Block[C-1].append(D-1)
    Block[D-1].append(C-1)

for i in range(N):
    ans = Friend_Net.size(i) - friend[i]
    for b in Block[i]:
        if Friend_Net.same(i, b):
            ans -= 1
    print(ans-1)

Solution (Python):
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 = map(int,input().split())
Friend_Net = UnionFind(N)
Friend = [[] for i in range(N)]
Block = [[] for i in range(N)]
friend = [0] * N

for i in range(M):
    A, B = map(int, input().split())
    friend[A-1] += 1
    friend[B-1] += 1
    Friend[A-1].append(B-1)
    Friend[B-1].append(A-1)
    Friend_Net.union(A-1, B-1)

for i in range(K):
    C, D = map(int, input().split())
    Block[C-1].append(D-1)
    Block[D-1].append(C-1)

for i in range(N):
    ans = Friend_Net.size(i) - friend[i]
    for b in Block[i]:
        if Friend_Net.same(i, b):
            ans -= 1
    print(ans-1)

Solution (Python):
n, m, k = map(int, input().split())
a_m, b_m = [], []
c_k, d_k = [], []
def minus(x):
    return int(x)-1

friends_list = [[] for i in range(n)]
block_list = [[] for i in range(n)]
for i in range(m):
    a_m.append(0)
    b_m.append(0)
    a_m[i], b_m[i] = map(minus, input().split())
    friends_list[a_m[i]].append(b_m[i])
    friends_list[b_m[i]].append(a_m[i])
for i in range(k):
    c_k.append(0)
    d_k.append(0)
    c_k[i], d_k[i] = map(minus, input().split())
    block_list[c_k[i]].append(d_k[i])
    block_list[d_k[i]].append(c_k[i])


def find(x):
    if parents[x] < 0:
        return x
    else:
        return find(parents[x])

def union(x, y):
    x = find(x)
    y = find(y)
    if x == y:
        return 0
    if parents[x] > parents[y]:
        x, y = y, x
    parents[x] += parents[y]
    parents[y] = x
    return 0

def union_size(x):
    return -parents[find(x)]

def blocks_in_union(x):
    count = 0
    for block in block_list[x]:
        if find(x) == find(block):
            count += 1
    return count

parents =[ -1 for i in range(n)]
for i in range(m):
    union(a_m[i], b_m[i])
for i in range(n):
    print(union_size(i)-1-len(friends_list[i])-blocks_in_union(i), end=' ')

Solution (Python):
import sys
sys.setrecursionlimit(10**5)
N,M,K = list(map(int,input().split(' ')))

friend_list = [0]*(N+1)
bad_list = [0]*(N+1)
group = {}

parent = [i for i in range(N+1)]
rank = [0]*(N+1)

def union(x,y):
    x = find(x)
    y = find(y)
    if x==y:
        return
    if rank[x]<rank[y]:
        parent[x]=y
    else:
        parent[y]=x
        if rank[x]==rank[y]:
            rank[x]+=1

def find(x):
    if parent[x] == x:
        return x
    else:
        parent[x]= find(parent[x])
        return parent[x]





for i in range(M):
    A,B = list(map(int,input().split(' ')))
    friend_list[A]+=1
    friend_list[B]+=1
    
    union(A,B)

for i in range(1,N+1):
    oya = find(i)
    if oya in group:
        group[oya] = group[oya]+1
    else:
        group[oya]=1

for i in range(K):
    A,B = list(map(int,input().split(' ')))
    if find(A)==find(B):
        bad_list[A]+=1
        bad_list[B]+=1


kouho = [0]*N

for i in range(1,N+1):
    oya = find(i)
    abc = group[oya]-friend_list[i]-bad_list[i]-1
    kouho[i-1] = abc

for k in kouho:
    print(k,end=" ")
print()

Solution (Python):
N, M, K = map(int, input().split())
F = [[] for _ in range(N)]
B = [[] for _ in range(N)]

for _ in range(M):
    a, b = map(int, input().split())
    a, b = a - 1, b - 1
    F[a].append(b)
    F[b].append(a)

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


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

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

    
    def union(self, x, y):
        x = self.find(x)
        y = self.find(y)

        if x == y:
            return

        if self.rank[x] < self.rank[y]:
            self.par[x] = y
            self.size[y] += self.size[x]
        else:
            self.par[y] = x
            self.size[x] += self.size[y]
            if self.rank[x] == self.rank[y]:
                self.rank[x] += 1

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

    
    def all_find(self):
        for n in range(len(self.par)):
            self.find(n)


UF = UnionFind(N)
for iam in range(N):
    for friend in F[iam]:
        UF.union(iam, friend)

ans = [UF.size[UF.find(iam)] - 1 for iam in range(N)]
for iam in range(N):
    ans[iam] -= len(F[iam])

for iam in range(N):
    for block in B[iam]:
        if UF.same(iam, block):
            ans[iam] -= 1

print(*ans, sep=' ')

Solution (Python):
class UnionFind:
    def __init__(self, n):
        self.n = n
        self.parent = [i for i in range(n)]  
        self.rank = [1] * n  
        self.size = [1] * n  

    def find(self, x):  
        if self.parent[x] == x:
            return x
        else:
            self.parent[x] = self.find(self.parent[x])  
            return self.parent[x]

    def unite(self, x, y):  
        x = self.find(x)
        y = self.find(y)
        if x != y:
            if self.rank[x] < self.rank[y]:
                self.parent[x] = y
                self.size[y] += self.size[x]
            else:
                self.parent[y] = x
                self.size[x] += self.size[y]
                if self.rank[x] == self.rank[y]:
                    self.rank[x] += 1

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

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


N, M, K = map(int, input().split())
A, B = [0] * M, [0] * M  
C, D = [0] * K, [0] * K  

direct =[[] for _ in range(N)]  
uf = UnionFind(N)  

for i in range(M):
    A[i], B[i] = map(int, input().split())
    A[i] -= 1
    B[i] -= 1
    direct[A[i]].append(B[i])
    direct[B[i]].append(A[i])
    uf.unite(A[i], B[i])

for i in range(K):
    C[i], D[i] = map(int, input().split())
    C[i] -= 1
    D[i] -= 1
    if uf.is_same(C[i], D[i]):
        direct[C[i]].append(D[i])
        direct[D[i]].append(C[i])

ans = [0] * N
for i in range(N):
    ans[i] = uf.group_size(i) - len(direct[i]) - 1
print(*ans)

Solution (Python):

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

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

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

    
    def same_check(self, x, y):
        return self.find(x) == self.find(y)
                
    
    def groupsize(self):
        cnt = 0
        for i,x in enumerate(self.par):
            if x == i:
                cnt += 1
        return cnt

    
    def size(self, x):
        return -self.root[self.find(x)]
    
    
    def members(self, x):
        ret = []
        rootNode = self.find(x)
        for i in range(len(self.par)):
            if self.find(i) == rootNode:
                ret.append(i)
        return ret
    
    
    def get_roots(self):
        ret = []
        for i,x in enumerate(self.par):
            if x==i:
                ret.append(i)
        return ret
    
    
    def all_members(self):
        L = len(self.par)
        roots = self.get_roots()
        namedict = [0]*L
        for i,r in enumerate(roots):
            namedict[r] = i
        
        
        ret = [[] for _ in range(len(roots))]
        for i in range(len(self.par)):
            p = namedict[self.find(i)]
            ret[p].append(i)
        return ret
            
    

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)]
    
uf = UnionFind(N)
for a,b in AB:
    uf.union(a-1,b-1)
    
ans = [0]*N
 
for group in uf.all_members():
    L = len(group)-1
    for a in group:
        ans[a] = L
        


for a,b in AB:
    ans[a-1] -= 1
    ans[b-1] -= 1
 
for c,d in CD:
    if uf.same_check(c-1,d-1):
        ans[c-1] -= 1
        ans[d-1] -= 1
 
print(*ans)

Solution (Python):

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

    
    def find(self, x):
        if self.par[x] == x:
            return x
        else:
            node = [x]
            cur = node[0]
            while self.par[cur] != cur:
                cur = self.par[cur]
                node.append(self.par[cur])
            
            for n in node:
                self.par[n] = cur
            
            return self.par[x]

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

    
    def same_check(self, x, y):
        return self.find(x) == self.find(y)
                
    
    def groupsize(self):
        cnt = 0
        for i,x in enumerate(self.par):
            if x == i:
                cnt += 1
        return cnt

    
    def size(self, x):
        return -self.root[self.find(x)]
    
    
    def members(self, x):
        ret = []
        rootNode = self.find(x)
        for i in range(len(self.par)):
            if self.find(i) == rootNode:
                ret.append(i)
        return ret
    
    
    def get_roots(self):
        ret = []
        for i,x in enumerate(self.par):
            if x==i:
                ret.append(i)
        return ret
    
    
    def all_members(self):
        L = len(self.par)
        roots = self.get_roots()
        namedict = [0]*L
        for i,r in enumerate(roots):
            namedict[r] = i
        
        
        ret = [[] for _ in range(len(roots))]
        for i in range(len(self.par)):
            p = namedict[self.find(i)]
            ret[p].append(i)
        return ret
            
    

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)]
    
uf = UnionFind(N)
for a,b in AB:
    uf.union(a-1,b-1)
    
ans = [0]*N
 
for group in uf.all_members():
    L = len(group)-1
    for a in group:
        ans[a] = L
        


for a,b in AB:
    ans[a-1] -= 1
    ans[b-1] -= 1
 
for c,d in CD:
    if uf.same_check(c-1,d-1):
        ans[c-1] -= 1
        ans[d-1] -= 1
 
print(*ans)

Solution (Python):

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

    
    def find(self, x):
        if self.par[x] == x:
            return x
        else:
            node = [x]
            cur = node[0]
            while self.par[cur] != cur:
                cur = self.par[cur]
                node.append(self.par[cur])
            
            for n in node:
                self.par[n] = cur
            
            return self.par[x]

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

    
    def same_check(self, x, y):
        return self.find(x) == self.find(y)
                
    
    def groupsize(self):
        cnt = 0
        for i,x in enumerate(self.par):
            if x == i:
                cnt += 1
        return cnt

    
    def size(self, x):
        return -self.root[self.find(x)]
    
    
    def members(self, x):
        ret = []
        rootNode = self.find(x)
        for i in range(len(self.par)):
            if self.find(i) == rootNode:
                ret.append(i)
        return ret
    
    
    def get_roots(self):
        ret = []
        for i,x in enumerate(self.par):
            if x==i:
                ret.append(i)
        return ret
    
    
    def all_members(self):
        roots = self.get_roots()
        namedict = {}
        for i,r in enumerate(roots):
            namedict[r] = i            
        ret = [[] for _ in range(len(roots))]
        for i in range(len(self.par)):
            p = namedict[self.find(i)]
            ret[p].append(i)
        return ret
            
    

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)]
    
uf = UnionFind(N)
for a,b in AB:
    uf.union(a-1,b-1)
    
ans = [0]*N
 
for group in uf.all_members():
    L = len(group)-1
    for a in group:
        ans[a] = L
        


for a,b in AB:
    ans[a-1] -= 1
    ans[b-1] -= 1
 
for c,d in CD:
    if uf.same_check(c-1,d-1):
        ans[c-1] -= 1
        ans[d-1] -= 1
 
print(*ans)

Solution (Python):

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

    
    def find(self, x):
        if self.par[x] == x:
            return x
        else:
            node = [x]
            cur = node[0]
            while self.par[cur] != cur:
                cur = self.par[cur]
                node.append(self.par[cur])
            
            for n in node:
                self.par[n] = cur
            
            return self.par[x]

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

    
    def same_check(self, x, y):
        return self.find(x) == self.find(y)
                
    
    def groupsize(self):
        cnt = 0
        for i,x in enumerate(self.par):
            if x == i:
                cnt += 1
        return cnt

    
    def size(self, x):
        return -self.root[self.find(x)]
    
    
    def members(self, x):
        ret = []
        rootNode = self.find(x)
        for i in range(len(self.par)):
            if self.find(i) == rootNode:
                ret.append(i)
        return ret
    
    
    def get_roots(self):
        ret = []
        for i,x in enumerate(self.par):
            if x==i:
                ret.append(i)
        return ret
    
    
    def all_members(self):
        roots = self.get_roots()
        namedict = {}
        for i,r in enumerate(roots):
            namedict[r] = i            
        ret = [[] for _ in range(len(roots))]
        for i in range(len(self.par)):
            p = namedict[self.find(i)]
            ret[p].append(i)
        return ret
            
    

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)]
    
uf = UnionFind(N)
for a,b in AB:
    uf.union(a-1,b-1)
    
ans = [0]*N
 
for group in uf.all_members():
    L = len(group)-1
    for a in group:
        ans[a] = L
        


for a,b in AB:
    ans[a-1] -= 1
    ans[b-1] -= 1
 
for c,d in CD:
    if uf.same_check(c-1,d-1):
        ans[c-1] -= 1
        ans[d-1] -= 1
 
print(*ans)

Solution (Python):
from collections import defaultdict


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_counts(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)
friends = defaultdict(int)
for _ in range(M):  
    a, b = map(lambda x: int(x)-1, input().split())
    uf.union(a, b)
    friends[a] += 1
    friends[b] += 1
blocked = defaultdict(list)
for _ in range(K):  
    c, d = map(lambda x: int(x)-1, input().split())
    blocked[c].append(d)
    blocked[d].append(c)
ans = []
for i in range(N):
    
    group_size = uf.size(i)
    
    i_frends = friends[i]
    
    cnt = 0
    for b in blocked[i]:
        if uf.same(i, b):
            cnt += 1
    ans.append(group_size - i_frends - cnt - 1)
print(" ".join(map(str, ans)))

Solution (Python):
class Uf:
    def __init__(self, N):
        self.p = list(range(N))
        self.rank = [0] * N
        self.size = [1] * N

    def root(self, x): 
        if self.p[x] != x:
            self.p[x] = self.root(self.p[x])

        return self.p[x]

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

    def unite(self, x, y): 
        u = self.root(x)
        v = self.root(y)

        if u == v: return

        if self.rank[u] < self.rank[v]:
            self.p[u] = v
            self.size[v] += self.size[u]
            self.size[u] = 0
        else:
            self.p[v] = u
            self.size[u] += self.size[v]
            self.size[v] = 0

            if self.rank[u] == self.rank[v]:
                self.rank[u] += 1

    def count(self, x): 
        return self.size[self.root(x)]

N,M,K=map(int,input().split())
A=[]
B=[]
C=[]
D=[]
for i in range(M):
    p,q=map(int,input().split())
    A.append(p-1)
    B.append(q-1)

for i in range(K):
    p,q=map(int,input().split())
    C.append(p-1)
    D.append(q-1)

Friend=Uf(N)
NG=[set() for _ in range(N)]

for i in range(M):
    Friend.unite(A[i],B[i])
    NG[A[i]].add(B[i])
    NG[B[i]].add(A[i])

for i in range(K):
    if Friend.same(C[i],D[i]): 
        NG[C[i]].add(D[i])
        NG[D[i]].add(C[i])

for i in range(N):
    print(Friend.count(i)-len(NG[i])-1)

Solution (Python):

import sys
sys.setrecursionlimit(500000)

class UF:
    
    def __init__(self,n):
        self.n = n
        self.parents = [-1] * (n+1) 
        self.rnks = [0] * (n+1)

    def find(self,n):
        if self.parents[n] < 0:
            return n
        else:
            self.parents[n] = self.find(self.parents[n])
            return self.parents[n]

    def union(self,a,b):
        a_par = self.find(a)
        b_par = self.find(b)

        if a_par == b_par:
            return
        if self.rnks[a_par] > self.rnks[b_par]:
            a_par,b_par = b_par,a_par
        if self.rnks[a_par] == self.rnks[b_par]:
            self.rnks[a_par] += 1
        self.parents[a_par] += self.parents[b_par]
        self.parents[b_par] = a_par

    def length(self,n):
        par = self.find(n)
        return abs(self.parents[par])

    def is_same(self,a,b):
        return self.find(a) == self.find(b)



def main():
    n,m,k = map(int,input().split())
    uf = UF(n)
    block_lst = [0 for _ in range(n+1)]
    friend_lst = [0 for _ in range(n+1)]
    
    for i in range(m):
        a,b = map(int,input().split())
        friend_lst[a] += 1
        friend_lst[b] += 1
        uf.union(a,b)
    

    
    for i in range(k):
        c,d = map(int,input().split())
        if uf.is_same(c,d):
            block_lst[c] += 1
            block_lst[d] += 1
    
    
    answer = []

    for i in range(1,n+1):
        f_o_f = uf.length(i)
        answer.append(f_o_f - friend_lst[i] - block_lst[i] - 1)


    print(" ".join(map(str,answer)))






if __name__ == '__main__':
    main()

Solution (Python):
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)
cnt=[0]*n

for _ in range(m):
    x,y=map(int,input().split())
    uf.union(x-1,y-1)
    cnt[x-1]+=1
    cnt[y-1]+=1
for _ in range(k):
    x,y=map(int,input().split())
    if uf.same(x-1,y-1):
        cnt[x-1]+=1
        cnt[y-1]+=1
    
for i in range(n):
    print(uf.size(i)-cnt[i]-1,end=" ")

Solution (Python):
class Uf:
    def __init__(self, N):
        self.p = list(range(N))
        self.rank = [0] * N
        self.size = [1] * N

    def root(self, x):
        if self.p[x] != x:
            self.p[x] = self.root(self.p[x])

        return self.p[x]

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

    def unite(self, x, y):
        u = self.root(x)
        v = self.root(y)

        if u == v: return

        if self.rank[u] < self.rank[v]:
            self.p[u] = v
            self.size[v] += self.size[u]
            self.size[u] = 0
        else:
            self.p[v] = u
            self.size[u] += self.size[v]
            self.size[v] = 0

            if self.rank[u] == self.rank[v]:
                self.rank[u] += 1

    def count(self, x):
        return self.size[self.root(x)]

N,M,K=map(int,input().split())
A=[]
B=[]
C=[]
D=[]
for i in range(M):
    p,q=map(int,input().split())
    A.append(p-1)
    B.append(q-1)

for i in range(K):
    p,q=map(int,input().split())
    C.append(p-1)
    D.append(q-1)

Friend=Uf(N)
NG=[set() for _ in range(N)]

for i in range(M):
    Friend.unite(A[i],B[i])
    NG[A[i]].add(B[i])
    NG[B[i]].add(A[i])

for i in range(K):
    if Friend.same(C[i],D[i]):
        NG[C[i]].add(D[i])
        NG[D[i]].add(C[i])

for i in range(N):
    print(Friend.count(i)-len(NG[i])-1)