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: Time Limit Exceeded)

use std::io;
use std::str::FromStr;
use std::collections::HashSet;

fn input_tuple() -> (usize, usize) {
    let mut line: String = String::new();
    io::stdin().read_line(&mut line);
    let vec: Vec<usize> = line.split_whitespace().map(|s| s.parse().unwrap()).collect();
    (vec[0], vec[1])
}

fn input_tuple_three() -> (usize, usize, usize) {
    let mut line: String = String::new();
    io::stdin().read_line(&mut line);
    let vec: Vec<usize> = line.split_whitespace().map(|s| s.parse().unwrap()).collect();
    (vec[0], vec[1], vec[2])
}

fn dfs(idx: usize, friends: &Vec<HashSet<usize>>, visited: &mut Vec<bool>, cluster: &mut HashSet<usize>) {
    // println!("{}, visited = {:?}", idx, visited);
    if visited[idx] {
        return;
    }
    cluster.insert(idx);
    visited[idx] = true;
    for i in &friends[idx] {
        dfs(*i, friends, visited, cluster);
    }
}

fn main() {
    let (n, m, k) = input_tuple_three();
    let mut friends: Vec<HashSet<usize>> = Vec::new();
    let mut blocks: Vec<HashSet<usize>> = Vec::new();
    for _ in 0..n {
        friends.push(HashSet::new());
    }
    for _ in 0..m {
        let (a, b) = input_tuple();
        friends[a - 1].insert(b - 1);
        friends[b - 1].insert(a - 1);
    }
    for _ in 0..n {
        blocks.push(HashSet::new());
    }
    for _ in 0..k {
        let (c, d) = input_tuple();
        blocks[c - 1].insert(d - 1);
        blocks[d - 1].insert(c - 1);
    }

    // println!("{:?}", friends);
    // println!("{:?}", blocks);
    let mut visited = vec![false; n];
    let mut clusters: Vec<HashSet<usize>> = Vec::new();
    for i in 0..n {
        if !visited[i] {
            let mut cluster: HashSet<usize> = HashSet::new();
            dfs(i, &friends, &mut visited, &mut cluster);
            clusters.push(cluster);
        }
    }

    let mut ans = vec![0usize; n];
    for cluster in &clusters {
        for c in cluster {
            let friends_num = cluster.intersection(&friends[*c]).count();
            let blocks_num = cluster.intersection(&blocks[*c]).count();
            ans[*c] = cluster.len() - friends_num - blocks_num - 1;
        }
    }
    for i in ans {
        print!("{} ", i);
    }
    println!();
}