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)

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)*}
    };

    ($next:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($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, [ $t:tt ]) => {
        {
            let len = read_value!($next, usize);
            (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:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

#[allow(unused_imports)]
use std::cmp::{min, max};

fn main() {
    input!{
        n: usize,
        m: usize,
        k: usize,
        es1: [(usize1, usize1); m],
        es2: [(usize1, usize1); k]
    }
    let mut deg = vec![1; n];
    let mut uf = UnionFind::new_points((0..n).collect::<Vec<usize>>());
    for &(a, b) in &es1 {
        uf.union(a, b);
        deg[a] += 1;
        deg[b] += 1;
    }
    let mut ans = vec![0; n];
    for i in 0..n {
        ans[i] = uf.size(&i) - deg[i];
    }
    for &(c, d) in &es2 {
        if uf.is_same(&c, &d) {
            ans[c] -= 1;
            ans[d] -= 1;
        }
    }
    print!("{}", ans[0]);
    for i in 1..n {
        print!(" {}", ans[i]);
    }
    println!("");
}

use std::collections::{HashSet, HashMap};

struct UFNode {
    parent: usize,
    rank: usize,
    size: usize
}

struct UnionFind<T> where T: std::cmp::Eq + std::hash::Hash {
    map: HashMap<T,usize>,
    nodes: Vec<UFNode>,
    next: usize,
    n_unions: usize
}

impl<T: std::cmp::Eq + std::hash::Hash> UnionFind<T> {
    fn new() -> UnionFind<T> {
        UnionFind{map: HashMap::new(), nodes: Vec::new(), next: 0, n_unions: 0}
    }

    fn new_points(v: Vec<T>) -> UnionFind<T> {
        let mut uf = UnionFind::new();
        for x in v {
            if uf.map.contains_key(&x) {
                continue;
            }
            uf.add_point(x);
        }
        uf
    }

    fn add_point(&mut self, x: T) -> usize {
        let id = self.next;
        self.map.insert(x,id);
        self.nodes.push(UFNode{parent: id, rank: 0, size: 1});
        self.next += 1;
        self.n_unions += 1;
        id
    }

    fn get_id(&self, x: &T) -> Option<usize> {
        self.map.get(x).map(|x| *x)
    }

    fn find(&mut self, id: usize) -> usize {
        let p = self.nodes[id].parent;
        if p == id {
            return id;
        }
        let pr = self.find(p);
        self.nodes[id].parent = pr;
        pr
    }

    fn union(&mut self, x: T, y: T) {
        let idx = self.get_id(&x).unwrap_or_else(|| self.add_point(x));
        let idy = self.get_id(&y).unwrap_or_else(|| self.add_point(y));
        let xp = self.find(idx);
        let yp = self.find(idy);
        if xp == yp {
            return;
        }
        let rx = self.nodes[xp].rank;
        let ry = self.nodes[yp].rank;
        if rx < ry {
            self.nodes[xp].parent = yp;
        } else if rx > ry {
            self.nodes[yp].parent = xp;
        } else {
            self.nodes[yp].parent = xp;
            self.nodes[xp].rank += 1;
        }
        let size = self.nodes[xp].size + self.nodes[yp].size;
        self.nodes[xp].size = size;
        self.nodes[yp].size = size;
        self.n_unions -= 1;
    }

    fn is_same(&mut self, x: &T, y: &T) -> bool {
        let idx = self.get_id(x);
        let idy = self.get_id(y);
        if idx == None || idy == None {
            false
        } else {
            self.find(idx.unwrap()) == self.find(idy.unwrap())
        }
    }

    fn n_points(&self) -> usize {
        self.next
    }

    fn n_unions(&self) -> usize {
        self.n_unions
    }

    fn group_id(&mut self, x: &T) -> usize {
        let id = self.get_id(x).unwrap();
        self.find(id)
    }

    fn size(&mut self, x: &T) -> usize {
        let id = self.get_id(x).unwrap();
        let idp = self.find(id);
        self.nodes[idp].size
    }

    fn to_sets(&mut self) -> HashMap<usize,HashSet<T>> where T: Clone {
        let mut map: HashMap<usize, HashSet<T>> = HashMap::new();
        for (k, v) in self.map.clone() {
            let mut s = map.entry(self.find(v)).or_insert_with(HashSet::new);
            s.insert(k);
        }
        map
    }
}