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 (1.42.0) (judge verdict: Accepted)

use proconio::{input, fastout};
use std::iter::*;
use itertools::Itertools;
pub struct UnionFind {
    n: usize,
    root: Vec<usize>,
    rank: Vec<usize>,
    size: Vec<usize>,
}

impl UnionFind {
    pub fn new(n: usize) -> Self {
        let root = (0..n).collect();
        let rank = vec![0;n];
        let size = vec![1;n];
        Self {
            n,
            root,
            rank,
            size,
        }
    }

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

    pub fn unite(&mut self, x: usize, y: usize) {
        if x >= self.n || y >= self.n {
            panic!();
        }
        let x_root = self.find(x);
        let y_root = self.find(y);
        if x_root == y_root {
            return;
        }

        //yに所属が移る
        if self.rank[x_root] < self.rank[y_root] {
            self.root[x_root] = y_root;
            self.size[y_root] += self.size[x_root];
        } else {
            //xに所属が移る
            self.root[y_root] = x_root;
            self.size[x_root] += self.size[y_root];
            //木の高さが同じとき，片方がもう片方のroot直下につながるので1だけ高くなる
            if self.rank[x_root] == self.rank[y_root] {
                self.rank[x_root] += 1;
            }
        }
    }

    pub fn size(&mut self, x: usize) -> usize {
        if x >= self.n {
            panic!();
        }
        let x_root = self.find(x);
        self.size[x_root]
    }

    pub fn is_same(&mut self, x: usize, y: usize) -> bool{
        if x >= self.n || y >= self.n {
            panic!();
        }
        self.find(x) == self.find(y)
    }

}

#[fastout]
fn main(){
    input!{
        n: usize,
        m: usize,
        k: usize,
        ab: [[usize; 2]; m],
        cd: [[usize; 2]; k],
    }

    let mut friend_relation = UnionFind::new(n);
    let mut friends = vec![0;n];
    for i in ab {
        friend_relation.unite(i[0]-1, i[1]-1);
        friends[i[0] - 1] += 1;
        friends[i[1] - 1] += 1;
    }
    let mut blocks = vec![Vec::new();n];
    for i in cd {
        blocks[i[0]-1].push(i[1]-1);
        blocks[i[1]-1].push(i[0]-1);
    }

    let res = (0..n)
              .map(|u| {
                  let all = friend_relation.size(u) - 1;
                  let mut res = all - friends[u];
                  for &b in &blocks[u] {
                      if friend_relation.is_same(u,b) {
                          res -= 1;
                      }
                  }
                  res
              })
              .join(" ");

    println!("{}", res);

}

// fn labeling() {

// }

// #[fastout]
// fn main() {
//     input!{
//         n: usize,
//         m: usize,
//         k: usize,
//         ab: [[usize; 2]; m],
//         cd: [[usize; 2]; k],
//     }

//     let mut friend: Vec<Vec<usize>> = vec![Vec::new();n];
//     let mut block: Vec<Vec<usize>> = vec![Vec::new();n];
//     let mut label: Vec<usize> = vec![n+1;n];
//     let mut label_size: Vec<usize> = vec![0;n];
//     for i in 0..m {
//         friend[ab[i][0] - 1].push(ab[i][1] - 1);
//         friend[ab[i][1] - 1].push(ab[i][0] - 1);
//         let mut new_label = min(ab[i][0]-1, ab[i][1]-1);
//         if label[ab[i][0]-1]!=n+1 || label[ab[i][1]-1]!=n+1 {
//             new_label = min(new_label, label[ab[i][0]-1]);
//             new_label = min(new_label, label[ab[i][1]-1]);
//         }
//         for j in 0..i+1 {
//             if label[j]==label[ab[i][0]-1] && label[j]==label[ab[j][1]-1]{
//                 label[j] = new_label;
//             }
//         }
//         label[ab[i][0]-1] = new_label;
//         label[ab[i][1]-1] = new_label;
//     }
//     for i in 0..n {
//         label_size[label[i]] += 1;
//     }
//     for i in 0..k {
//         if label[cd[i][0]-1]!=label[cd[i][1]-1] {
//             continue;
//         } else {
//             block[cd[i][0] - 1].push(cd[i][1] - 1);
//             block[cd[i][1] - 1].push(cd[i][0] - 1);
//         }
//     }

//     for i in 0..n {
//         //print!("{} ", label_size[i]);
//         // print!("{} ", label[i]);
//         print!("{} ", block[i].len());
//         // let ans = label_size[label[i]] - 1 - friend[i].len() - block[i].len();
//         // print!("{} ", ans);
//     }

//     print!("\n");
// }