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: Time Limit Exceeded)

// This code is generated by [cargo-atcoder](https://github.com/tanakh/cargo-atcoder)
// Original source code:
/*
#[allow(unused_macros)]
macro_rules! debug {
	($($a:expr),*) => {
		println!(concat!($(stringify!($a), " = {:?}, "),*), $($a),*);
	}
}

use std::cmp::{max, min};
#[allow(dead_code)]
const MOD: usize = 1e9 as usize + 7;

pub struct UnionFind(Vec<usize>);

impl UnionFind {
    pub fn new(n: usize) -> UnionFind {
        UnionFind((0..n).collect())
    }

    pub fn find(&mut self, i: usize) -> usize {
        if self.0[i] == i {
            i
        } else {
            let p = self.0[i];
            self.0[i] = self.find(p);
            self.0[i]
        }
    }

    pub fn union(&mut self, i: usize, j: usize) {
        let ni = self.find(i);
        let nj = self.find(j);
        self.0[ni] = nj;
    }
}
fn main() {
    let (r, w) = (std::io::stdin(), std::io::stdout());
    let mut sc = IO::new(r.lock(), w.lock());

    let n: usize = sc.read();
    let m: usize = sc.read();
    let k: usize = sc.read();
    let ab = sc.read_pairs::<usize, usize>(m);
    let cd = sc.read_pairs::<usize, usize>(k);

    let mut friend = vec![vec![false; n]; n];
    for (a, b) in ab.iter() {
        friend[*a - 1][*b - 1] = true;
        friend[*b - 1][*a - 1] = true;
    }
    let mut block = vec![vec![false; n]; n];
    for (c, d) in cd.into_iter() {
        block[c - 1][d - 1] = true;
        block[d - 1][c - 1] = true;
    }

    let mut uf = UnionFind::new(n);
    for (a, b) in ab.into_iter() {
        uf.union(a - 1, b - 1);
    }

    for i in 0..n {
        let mut ans = 0;
        for j in 0..n {
            if i == j {
                continue;
            }
            if uf.find(i) == uf.find(j) && friend[i][j] == false && block[i][j] == false {
                ans += 1;
            }
        }
        print!("{}", ans);
        if i != n - 1 {
            print!(" ");
        } else {
            println!("");
        }
    }
}

pub struct IO<R, W: std::io::Write>(R, std::io::BufWriter<W>);

impl<R: std::io::Read, W: std::io::Write> IO<R, W> {
    pub fn new(r: R, w: W) -> IO<R, W> {
        IO(r, std::io::BufWriter::new(w))
    }
    pub fn write<S: std::ops::Deref<Target = str>>(&mut self, s: S) {
        use std::io::Write;
        self.1.write(s.as_bytes()).unwrap();
    }
    pub fn read<T: std::str::FromStr>(&mut self) -> T {
        use std::io::Read;
        let buf = self
            .0
            .by_ref()
            .bytes()
            .map(|b| b.unwrap())
            .skip_while(|&b| b == b' ' || b == b'\n' || b == b'\r' || b == b'\t')
            .take_while(|&b| b != b' ' && b != b'\n' && b != b'\r' && b != b'\t')
            .collect::<Vec<_>>();
        unsafe { std::str::from_utf8_unchecked(&buf) }
            .parse()
            .ok()
            .expect("Parse error.")
    }
    pub fn read_vec<T: std::str::FromStr>(&mut self, n: usize) -> Vec<T> {
        (0..n).map(|_| self.read()).collect()
    }
    pub fn read_chars(&mut self) -> Vec<char> {
        self.read::<String>().chars().collect()
    }
    pub fn read_char_grid(&mut self, n: usize) -> Vec<Vec<char>> {
        (0..n).map(|_| self.read_chars()).collect()
    }
    pub fn read_pairs<T: std::str::FromStr, U: std::str::FromStr>(
        &mut self,
        n: usize,
    ) -> Vec<(T, U)> {
        (0..n)
            .map(|_| (self.read::<T>(), self.read::<U>()))
            .collect::<Vec<(T, U)>>()
    }
    pub fn read_matrix<T: std::str::FromStr>(&mut self, n: usize, m: usize) -> Vec<Vec<T>> {
        (0..n)
            .map(|_| (0..m).map(|_| self.read()).collect())
            .collect()
    }
}
*/

fn main() {
    let exe = "/tmp/binFBFC0914";
    std::io::Write::write_all(&mut std::fs::File::create(exe).unwrap(), &decode(BIN)).unwrap();
    std::fs::set_permissions(exe, std::os::unix::fs::PermissionsExt::from_mode(0o755)).unwrap();
    std::process::exit(std::process::Command::new(exe).status().unwrap().code().unwrap())
}

fn decode(v: &str) -> Vec<u8> {
    let mut ret = vec![];
    let mut buf = 0;
    let mut tbl = vec![64; 256];
    for i in 0..64 { tbl[TBL[i] as usize] = i as u8; }
    for (i, c) in v.bytes().filter_map(|c| { let c = tbl[c as usize]; if c < 64 { Some(c) } else { None } }).enumerate() {
        match i % 4 {
            0 => buf = c << 2,
            1 => { ret.push(buf | c >> 4); buf = c << 4; }
            2 => { ret.push(buf | c >> 2); buf = c << 6; }
            3 => ret.push(buf | c),
            _ => unreachable!(),
        }
    }
    ret
}

const TBL: &'static [u8] = b"ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
const BIN: &'static str = "<base64-encoded ELF executable removed>";