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 (1.42.0) (judge verdict: Accepted)

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

use proconio::{input, fastout};
use proconio::marker::*;
use whiteread::parse_line;
use std::collections::*;
use num::*;
use num_traits::*;
use superslice::*;
use std::ops::*;

use itertools::Itertools;
use itertools_num::ItertoolsNum;

#[derive(Debug, Clone)]
pub struct UnionFind<K> {
    // For element at index *i*, store the index of its parent; the representative itself
    // stores its own index. This forms equivalence classes which are the disjoint sets, each
    // with a unique representative.
    parent: Vec<K>,

    // size
    size: Vec<usize>,
}

#[inline]
unsafe fn get_unchecked<K>(xs: &[K], index: usize) -> &K {
    debug_assert!(index < xs.len());
    xs.get_unchecked(index)
}

#[inline]
unsafe fn get_unchecked_mut<K>(xs: &mut [K], index: usize) -> &mut K {
    debug_assert!(index < xs.len());
    xs.get_unchecked_mut(index)
}

impl<K> UnionFind<K>
where
    K: PrimInt,
{
    /// Create a new `UnionFind` of `n` disjoint sets.
    pub fn new(n: usize) -> Self {
        let size = vec![1; n];
        let parent = (0..n).map(|x| K::from(x).unwrap()).collect::<Vec<K>>();

        UnionFind { parent, size }
    }

    /// Return the representative for `x`.
    ///
    /// **Panics** if `x` is out of bounds.
    pub fn find(&self, x: K) -> K {
        assert!(x.to_usize().unwrap() < self.parent.len());
        unsafe {
            let mut x = x;
            loop {
                // Use unchecked indexing because we can trust the internal set ids.
                let xparent = *get_unchecked(&self.parent, x.to_usize().unwrap());
                if xparent == x {
                    break;
                }
                x = xparent;
            }
            x
        }
    }

    /// Return the representative for `x`.
    ///
    /// Write back the found representative, flattening the internal
    /// datastructure in the process and quicken future lookups.
    ///
    /// **Panics** if `x` is out of bounds.
    pub fn find_mut(&mut self, x: K) -> K {
        assert!(x.to_usize().unwrap() < self.parent.len());
        unsafe { self.find_mut_recursive(x) }
    }

    unsafe fn find_mut_recursive(&mut self, mut x: K) -> K {
        let mut parent = *get_unchecked(&self.parent, x.to_usize().unwrap());
        while parent != x {
            let grandparent = *get_unchecked(&self.parent, parent.to_usize().unwrap());
            *get_unchecked_mut(&mut self.parent, x.to_usize().unwrap()) = grandparent;
            x = parent;
            parent = grandparent;
        }
        x
    }

    /// Returns `true` if the given elements belong to the same set, and returns
    /// `false` otherwise.
    pub fn equiv(&self, x: K, y: K) -> bool {
        self.find(x) == self.find(y)
    }

    /// Unify the two sets containing `x` and `y`.
    ///
    /// Return `false` if the sets were already the same, `true` if they were unified.
    ///
    /// **Panics** if `x` or `y` is out of bounds.
    pub fn union(&mut self, x: K, y: K) -> bool {
        if x == y {
            return false;
        }
        let xrep = self.find_mut(x);
        let yrep = self.find_mut(y);

        if xrep == yrep {
            return false;
        }

        let xrepu = xrep.to_usize().unwrap();
        let yrepu = yrep.to_usize().unwrap();
        let xsize = self.size[xrepu];
        let ysize = self.size[yrepu];

        // The rank corresponds roughly to the depth of the treeset, so put the
        // smaller set below the larger
        if xsize > ysize {
            self.parent[yrepu] = xrep;
            self.size[xrepu] += ysize;
        } else {
            self.parent[xrepu] = yrep;
            self.size[yrepu] += xsize;
        }
        
        true
    }

    /// Return a vector mapping each element to its representative.
    pub fn into_labeling(mut self) -> Vec<K> {
        // write in the labeling of each element
        unsafe {
            for ix in 0..self.parent.len() {
                let k = *get_unchecked(&self.parent, ix);
                let xrep = self.find_mut_recursive(k);
                *self.parent.get_unchecked_mut(ix) = xrep;
            }
        }
        self.parent
    }

    pub fn size(&self, x: K) -> usize {
        let xrep = self.find(x);
        let xrepu = xrep.to_usize().unwrap();

        self.size[xrepu]
    }
}

#[fastout]
fn solve() {
    const MOD: usize = 1_000_000_007;
    const INF: usize = std::usize::MAX;

    input!{
        n: usize, m: usize, k: usize,
        ab_vec: [(usize, usize); m],
        cd_vec: [(usize, usize); k],
    }

    let mut friend_un: UnionFind<usize> = UnionFind::new(n);

    // let friend_set: HashSet<(usize, usize)> = ab_vec.into_iter().collect();

    let mut friend_vec = vec![0; n];

    for (a, b) in ab_vec.iter() {
        friend_vec[a-1] += 1;
        friend_vec[b-1] += 1; 
        friend_un.union(a-1, b-1);
    }

    let mut block_vec = vec![vec![]; n];
    for (c, d) in cd_vec.iter() {
        block_vec[c-1].push(d-1);
        block_vec[d-1].push(c-1);
    }

    for i in 0..n {
        let all = friend_un.size(i);
        let mut ans = all - friend_vec[i] - 1;

        for &b in block_vec[i].iter() {
            if friend_un.equiv(i, b) {
                ans -= 1;
            }
        }

        println!("{}", ans)
    }
}

fn main() {
    solve();
}