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::str::FromStr;

fn get_one<T : FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    return s.trim().parse().ok().unwrap();
}

fn get_vec<T : FromStr>() -> Vec<T> {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    return s.trim().split_whitespace().map(|e| e.parse().ok().unwrap()).collect();
}

fn dfs(edges : &Vec<Vec<u32>>, mut visited : &mut Vec<i32>, cur : u32, mark : u32) -> i32 {
    visited[cur as usize] = mark as i32;
    let mut cnt = 0;
    for next in &edges[cur as usize] {
        if visited[*next as usize] == -1 {
            let x = dfs(&edges, &mut visited, *next, mark);
            cnt += x;
        }
    }
    return cnt+1;
}

fn main() {
    let nmk : Vec<usize> = get_vec();
    let mut edges : Vec<Vec<u32>> = vec!(Vec::new();nmk[0]);
    let mut ecnt = vec!(0;nmk[0]);
    for i in 0..nmk[1] {
        let ab : Vec<u32> = get_vec();
        edges[ab[0] as usize -1].push(ab[1]-1);
        ecnt[ab[0] as usize -1] +=1;
        ecnt[ab[1] as usize -1] +=1;
        edges[ab[1] as usize -1].push(ab[0]-1);
    }

    let mut visited : Vec<i32> = vec!(-1;nmk[0]);
    let mut cnt : Vec<i32> = vec!(0;nmk[0]);
    for i in 0..nmk[0] {
        if visited[i] == -1 {
            let ret = dfs(&edges, &mut visited, i as u32, i as u32);
            cnt[i] = ret;
        }
        cnt[i] = cnt[visited[i] as usize];
    }
    let mut blocks : Vec<i32> = vec!(0;nmk[0]);
    for i in 0..nmk[2] {
        let cd : Vec<u32> = get_vec();
        if visited[cd[0] as usize -1 as usize] == visited[cd[1] as usize -1 as usize] {
            blocks[cd[0] as usize -1] += 1;
            blocks[cd[1] as usize -1] += 1;
        }
    }
    let mut s = String::new();
    for i in 0..nmk[0]{
        let x = cnt[i] - ecnt[i] as i32 - blocks[i] -1;
        s = format!("{} {}", s, x.to_string());
    }
    println!("{}", s.trim());
}