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)

//! ----------------------------------------------
//! Framework <https://github.com/vain0x/procon>
//!
//! See the bottom of file for solution.
//! ----------------------------------------------

#![allow(unused_imports)]
#![allow(non_snake_case)]

use std::cell::RefCell;
use std::cmp::{max, min, Ordering};
use std::collections::*;
use std::fmt::{Debug, Display, Formatter, Write as FmtWrite};
use std::io::{stderr, stdin, BufRead, Write};
use std::mem::{replace, swap};
use std::ops::*;
use std::rc::Rc;

/// Print values to standard error if debug mode.
#[allow(unused_macros)]
macro_rules! debug {
    ($($e:expr),*) => {
        #[cfg(debug_assertions)]
        $({
            let (e, mut err) = (stringify!($e), stderr());
            writeln!(err, "\x1B[33m{}\x1B[0m = {:?}", e, $e).unwrap()
        })*
    };
}

/// Read from standard input and parse each word.
/// - `read!(T, U, ..)` parses a line as a tuple of words.
/// - `read![[T]]` parses a line as an array of words.
/// - `read![..; N]` parses `N` lines, using `read!(..)` repeatedly.
#[allow(unused_macros)]
macro_rules! read {
    ([$t:ty] ; $n:expr) =>
        ((0..$n).map(|_| read!([$t])).collect::<Vec<_>>());
    ($($t:ty),+ ; $n:expr) =>
        ((0..$n).map(|_| read!($($t),+)).collect::<Vec<_>>());
    ([$t:ty]) =>
        (rl().split_whitespace().map(|w| w.parse().unwrap()).collect::<Vec<$t>>());
    ($($t:ty),*) => {{
        let buf = rl();
        let mut w = buf.split_whitespace();
        ($(w.next().unwrap().parse::<$t>().unwrap()),*)
    }};
}

/// Read a line from standard input.
#[allow(dead_code)]
fn rl() -> String {
    let mut buf = String::new();
    stdin().read_line(&mut buf).unwrap();

    #[allow(deprecated)]
    buf.trim_right().to_owned()
}

// -----------------------------------------------
// Solution
// -----------------------------------------------

struct Solver {
    N: usize,
    G: Vec<Vec<usize>>,
    root: Vec<usize>,
    size: Vec<usize>,
    friendship: Vec<(usize, usize)>,
    block: Vec<(usize, usize)>,
}

impl Solver {
    fn new(N: usize, friendship: Vec<(usize, usize)>, block: Vec<(usize, usize)>) -> Solver {
        let mut G = vec![vec![]; N];

        for &(u, v) in &friendship {
            G[u].push(v);
            G[v].push(u);
        }

        Solver {
            N: N,
            G: G,
            root: vec![N; N],
            size: vec![0; N],
            friendship: friendship,
            block: block,
        }
    }

    fn dfs(&mut self, u: usize, root: usize) {
        if self.root[u] != self.N {
            return;
        }

        self.root[u] = root;
        self.size[root] += 1;

        for i in 0..self.G[u].len() {
            let v = self.G[u][i];

            self.dfs(v, root);
        }
    }

    fn solve(&mut self) {
        let N = self.N;

        for u in 0..N {
            self.dfs(u, u);
        }

        let mut friend_count = vec![0; N];
        let mut block_count = vec![0; N];

        for &(u, v) in &self.friendship {
            if self.root[u] == self.root[v] {
                friend_count[u] += 1;
                friend_count[v] += 1;
            }
        }

        for &(u, v) in &self.block {
            if self.root[u] == self.root[v] {
                block_count[u] += 1;
                block_count[v] += 1;
            }
        }

        let mut candidate_count = vec![0; N];
        for u in 0..N {
            candidate_count[u] = self.size[self.root[u]] - friend_count[u] - block_count[u] - 1;
        }

        println!(
            "{}",
            (0..N)
                .map(|u| candidate_count[u].to_string())
                .collect::<Vec<_>>()
                .join(" ")
        );
    }
}

fn main() {
    let (N, M, K) = read!(usize, usize, usize);

    let mut friendship = vec![];
    for _ in 0..M {
        let (u, v) = read!(usize, usize);
        friendship.push((u - 1, v - 1));
    }

    let mut block = vec![];
    for _ in 0..K {
        let (u, v) = read!(usize, usize);
        block.push((u - 1, v - 1));
    }

    let mut solver = Solver::new(N, friendship, block);
    solver.solve();
}