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 in Rust (1.15.1) (judge verdict: Accepted)

use std::cmp;
use std::collections::HashMap;
use std::collections::VecDeque;
use std::usize::MAX;

macro_rules! input {(source = $s:expr, $($r:tt)*) => {let mut iter = $s.split_whitespace();let mut next = || { iter.next().unwrap() };input_inner!{next, $($r)*}};($($r:tt)*) => {let stdin = std::io::stdin();let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));let mut next = move || -> String{bytes.by_ref().map(|r|r.unwrap() as char).skip_while(|c|c.is_whitespace()).take_while(|c|!c.is_whitespace()).collect()};input_inner!{next, $($r)*}};}
macro_rules! input_inner {($next:expr) => {};($next:expr, ) => {};($next:expr, $var:ident : $t:tt $($r:tt)*) => {let $var = read_value!($next, $t);input_inner!{$next $($r)*}};}
macro_rules! read_value {($next:expr, ( $($t:tt),* )) => {( $(read_value!($next, $t)),* )};($next:expr, [ $t:tt ; $len:expr ]) => {(0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()};($next:expr, chars) => {read_value!($next, String).chars().collect::<Vec<char>>()};($next:expr, usize1) => {read_value!($next, usize) - 1};($next:expr, [ $t:tt ]) => {{let len = read_value!($next, usize);(0..len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()}};($next:expr, $t:ty) => {$next().parse::<$t>().expect("Parse error")};}

fn solve() {
    input! {
        n: usize,
        m: usize,
        k: usize,
        friend: [(usize1,usize1);m],
        block: [(usize1,usize1);k]
    }

    let mut friend_count = vec![0; n];
    let mut direct_blocks = vec![vec![]; n];

    let mut uf = UnionFind::new(n);
    for f in &friend {
        uf.unite(f.0, f.1);
        friend_count[f.0] += 1;
        friend_count[f.1] += 1;
    }

    for b in &block {
        direct_blocks[b.0].push(b.1);
        direct_blocks[b.1].push(b.0);
    }

    for i in 0..n {
        let mut candidate = uf.size(i) - friend_count[i] - 1;
        for &b in &direct_blocks[i] {
            if uf.same(i, b) {
                candidate -= 1;
            }
        }
        if i != n - 1 {
            print!("{} ", candidate);
        } else {
            print!("{}", candidate);
        }
    }
    println!("");
}

fn main() {
    let thd = std::thread::Builder::new().stack_size(104_857_600);
    thd.spawn(|| solve()).unwrap().join().unwrap();
}

struct UnionFind {
    parent: Vec<usize>,
    rank: Vec<usize>,
    size: Vec<usize>,
}

impl UnionFind {
    fn new(n: usize) -> UnionFind {
        UnionFind {
            parent: (0..n).collect(),
            rank: vec![0; n],
            size: vec![1; n],
        }
    }

    fn root(&mut self, x: usize) -> usize {
        let parent = self.parent[x];
        if parent != x {
            self.parent[x] = self.root(parent);
            self.rank[x] = 1;
        }
        return self.parent[x];
    }

    fn unite(&mut self, x: usize, y: usize) {
        if self.same(x, y) {
            return;
        }
        let x_root = self.root(x);
        let y_root = self.root(y);
        if self.rank[x_root] >= self.rank[y_root] {
            self.size[x_root] += self.size(y);
            self.rank[y_root] = 1;
            self.parent[y_root] = x_root;
        } else {
            let x_root = self.root(x);
            let y_root = self.root(y);
            self.size[y_root] += self.size(x);
            self.rank[x_root] = 1;
            self.parent[x_root] = y_root;
        }
    }

    fn size(&mut self, x: usize) -> usize {
        let x_root = self.root(x);
        return self.size[x_root];
    }

    fn same(&mut self, x: usize, y: usize) -> bool {
        self.root(x) == self.root(y)
    }
}

const LARGE_PRIME: u64 = 1_000_000_007;

// @〜でsnippet