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)

use std::io::*;
use std::str::FromStr;

pub fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

pub fn read_n<T: FromStr>(n: usize) -> Vec<T> {
    read_n_logic::<T>(n, vec![])
}

pub fn read_n_logic<T: FromStr>(n: usize, mut a: Vec<T>) -> Vec<T> {
    match n {
        0 => a,
        _ => {
            a.push(read::<T>());
            read_n_logic(n - 1, a)
        }
    }
}
use std::cmp::{max, min};
use std::collections::HashMap;

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

impl UnionFind {
    fn new(n: usize) -> UnionFind {
        // initialize
        let mut v: Vec<usize> = vec![];
        for i in 0..n + 1 {
            v.push(i)
        }
        let mut s: Vec<usize> = vec![1; n + 1];

        UnionFind { parent: v, size: s }
    }

    fn root(&self, x: usize) -> usize {
        if self.parent[x] == x {
            return x;
        } else {
            return self.root(self.parent[x]);
        }
    }

    fn size(&self, a: usize) -> usize {
        let ra = self.root(a);
        self.size[ra]
    }

    fn unite(&mut self, a: usize, b: usize) {
        let mut ra = self.root(a);
        let mut rb = self.root(b);
        if ra == rb {
            return;
        }
        if self.size[ra] < self.size[rb] {
            std::mem::swap(&mut ra, &mut rb)
        };
        self.size[ra] += self.size[rb];
        self.parent[rb] = ra;
    }

    fn same(&self, a: usize, b: usize) -> bool {
        let ra = self.root(a);
        let rb = self.root(b);
        ra == rb
    }
}

fn main() {
    let n: usize = read();
    let m: usize = read();
    let k: usize = read();
    let mut map: HashMap<i64, Vec<i64>> = HashMap::new();
    let mut map_anchi: HashMap<i64, Vec<i64>> = HashMap::new();
    let mut uf = UnionFind::new(n);
    for _ in 0..m {
        let a: i64 = read();
        let b: i64 = read();
        uf.unite(a as usize, b as usize);
        hashmap_process(&mut map, a, b);
    }

    for _ in 0..k {
        let a: i64 = read();
        let b: i64 = read();
        hashmap_process(&mut map_anchi, a, b);
    }

    for i in 1..n + 1 {
        let i_i64 = i as i64;
        let mut ans = uf.size(i) - 1;
        ans -= match map.get(&i_i64) {
            None => 0,
            Some(s) => s.len(),
        };
        ans -= match map_anchi.get(&i_i64) {
            None => 0,
            Some(s) => {
                let mut cnt = 0;
                for j in s {
                    if uf.same(i as usize, *j as usize) {
                        cnt += 1
                    }
                }
                cnt
            }
        };
        print!("{} ", ans);
    }
}

fn hashmap_process(map: &mut HashMap<i64, Vec<i64>>, a: i64, b: i64) {
    if map.contains_key(&a) {
        let current = map.get_mut(&a).unwrap();
        current.push(b);
    } else {
        map.insert(a, vec![b]);
    }

    if map.contains_key(&b) {
        let current = map.get_mut(&b).unwrap();
        current.push(a);
    } else {
        map.insert(b, vec![a]);
    }
}