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)

//---------- begin union_find ----------
#[allow(dead_code)]
mod union_find {
    pub struct UF {
        p: Vec<i32>,
    }
    impl UF {
        pub fn new(n: usize) -> UF {
            UF {p: vec![-1; n] }
        }
        pub fn init(&mut self) {
            for p in self.p.iter_mut() {
                *p = -1;
            }
        }
        pub fn root(&self, mut x: usize) -> usize {
            while self.p[x] >= 0 {
                x = self.p[x] as usize;
            }
            x
        }
        pub fn same(&self, x: usize, y: usize) -> bool {
            self.root(x) == self.root(y)
        }
        pub fn unite(&mut self, mut x: usize, mut y: usize) -> Option<(usize, usize)> {
            x = self.root(x);
            y = self.root(y);
            if x == y {
                return None;
            }
            if self.p[x] > self.p[y] {
                let s = x;
                x = y;
                y = s;
            }
            self.p[x] += self.p[y];
            self.p[y] = x as i32;
            Some((x, y))
        }
        pub fn get_size(&self, x: usize) -> usize {
            let r = self.root(x);
            (-self.p[r]) as usize
        }
    }
}
//---------- end union_find ----------
//https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8 より
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};
    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
}

macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };
    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };
    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };
    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };
    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}

//

fn run() {
    input! {
        n: usize,
        m: usize,
        k: usize,
        a: [(usize1, usize1); m],
        b: [(usize1, usize1); k],
    }
    let mut friend = union_find::UF::new(n);
    let mut cnt = vec![0; n];
    for (a, b) in a {
        friend.unite(a, b);
        cnt[a] += 1;
        cnt[b] += 1;
    }
    let mut block = vec![vec![]; n];
    for (a, b) in b {
        block[a].push(b);
        block[b].push(a);
    }
    let mut out = String::new();
    for (i, (block, cnt)) in block.iter().zip(cnt.iter()).enumerate() {
        let mut ans = friend.get_size(i) - *cnt - 1;
        for b in block.iter() {
            if friend.same(i, *b) {
                ans -= 1;
            }
        }
        out.push_str(&format!("{} ", ans));
    }
    out.pop();
    println!("{}", out);
}

fn main() {
    run();
}