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)

// This code is generated by [cargo-atcoder](https://github.com/tanakh/cargo-atcoder)
// Original source code:
/*
use proconio::input;

fn main() {
    input! {
        n: usize,
        m: usize,
        k: usize,
        friends: [(usize, usize); m],
        blocks: [(usize, usize); k],
    }

    // eprintln!("{:?}", friends);
    // eprintln!("{:?}", blocks);

    let mut uf = UnionFind::new(n);

    for &(f1, f2) in &friends {
        uf.unite(f1 - 1, f2 - 1);
    }

    let mut friends_count = vec![0; n];
    for &(f1, f2) in &friends {
        friends_count[f1 - 1] += 1;
        friends_count[f2 - 1] += 1;
    }

    let mut blocks_count = vec![0; n];
    for &(b1, b2) in &blocks {
        if uf.same(b1 - 1, b2 - 1) {
            blocks_count[b1 - 1] += 1;
            blocks_count[b2 - 1] += 1;
        }
    }

    let mut count_roots = vec![0; n];
    for i in 0..n {
        let root = uf.find(i);
        count_roots[root] += 1;
    }

    let mut ans = vec![];
    for i in 0..n {
        let count = count_roots[uf.find(i)] - friends_count[i] - blocks_count[i] - 1;
        ans.push(count);
    }

    let ans_s = ans.iter().map(|a| a.to_string()).collect::<Vec<_>>().join(" ");
    println!("{}", ans_s);
}

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

impl UnionFind {
    fn new(n: usize) -> UnionFind {
        let mut parent = vec![];
        for i in 0..n {
            parent.push(i);
        }
        UnionFind {
            parent: parent,
            rank: vec![0; n],
        }
    }

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

    fn unite(&mut self, i: usize, j: usize) {
        let i = self.find(i);
        let j = self.find(j);
        if i == j {
            return;
        }

        if self.rank[i] < self.rank[j] {
            self.parent[i] = j;
        } else {
            self.parent[j] = i;
            if self.rank[i] == self.rank[j] {
                self.rank[i] += 1;
            }
        }
    }

    fn same(&mut self, i: usize, j: usize) -> bool {
        self.find(i) == self.find(j)
    }
}
*/

fn main() {
    let exe = "/tmp/bin8F3D9833";
    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>";