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)

#[allow(dead_code)]
fn is_whitespace(c: &char) -> bool {
    c.is_whitespace()
}

#[allow(dead_code)]
fn is_eol(c: &char) -> bool {
    ['\r', '\n'].contains(c)
}

#[allow(unused_macros)]
macro_rules! input {
    ($var:ident, $($r:tt)*) => {
        input_inner!{$var, $($r)*}
    };

    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        #[allow(unused_mut, unused_variables, bare_trait_objects)]
        let mut next = move |f: &Fn(&char)->bool| -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|f(&c))
                .take_while(|c|!f(&c))
                .collect()
        };
        input_inner!{next, $($r)*};
    };
}

#[allow(unused_macros)]
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)*}
    };
}

#[allow(unused_macros)]
macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; eol ]) => {
        $next(&is_eol).split(' ').map(|x| x.parse::<$t>().unwrap()).collect::<Vec<_>>()
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, bytes) => {
        read_value!($next, String).into_bytes()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, parser) => {
        $next
    };

    ($next:expr, $t:ty) => {
        $next(&is_whitespace).parse::<$t>().expect("Parse error")
    };
}

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

#[allow(dead_code)]
struct Edge {
    from: usize,
    to: usize,
}

#[allow(dead_code)]
type Graph = Vec<Vec<usize>>;
#[allow(dead_code)]
const M: usize = 1_000_000_007;

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

impl UnionFind {
    pub fn new(n: usize) -> UnionFind {
        let mut parent = vec![];
        for i in 0..n {
            parent.push(i);
        }
        UnionFind {
            parent: parent,
            size: vec![1; n],
            rank: vec![0; n],
        }
    }

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

    pub fn unite(&mut self, x: usize, y: usize) {
        let x_root = self.root(x);
        let y_root = self.root(y);

        if x_root == y_root {
            return;
        }

        let sum = self.size[x_root] + self.size[y_root];
        self.size[x_root] = sum;
        self.size[y_root] = sum;

        if self.rank[x_root] < self.rank[y_root] {
            self.parent[x_root] = y_root;
        } else {
            self.parent[y_root] = x_root;
            if self.rank[x_root] == self.rank[y_root] {
                self.rank[x_root] += 1;
            }
        }
    }

    pub fn same(&mut self, x: usize, y: usize) -> bool {
        self.root(x) == self.root(y)
    }

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

fn main() {
    input! {
        n: usize, m: usize, k: usize,
        ab: [(usize1, usize1); m],
        cd: [(usize1, usize1); k],
    }

    let mut uf = UnionFind::new(n);

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

    let mut graph = vec![vec![]; n];
    for (f, t) in cd {
        graph[f].push(t);
        graph[t].push(f);
    }

    for (f, t) in ab {
        uf.unite(f, t);
        numfriends[f] += 1;
        numfriends[t] += 1;
    }

    for i in 0..n {
        let mut cand = uf.size(i) - 1;
        cand -= numfriends[i];

        for &x in &graph[i] {
            if uf.same(x, i) {
                cand -= 1;
            }
        }
        if i != 0 {
            print!(" ");
        }
        print!("{}", cand);
    }
}