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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {

  static List<Connect> friend = new ArrayList<>();
  static List<Connect> block = new ArrayList<>();

  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int N = sc.nextInt();
    int M = sc.nextInt();
    int K = sc.nextInt();

    for (int i = 0; i < M; i++) {
      friend.add(new Connect(sc.nextInt(), sc.nextInt()));
    }
    for (int i = 0; i < K; i++) {
      block.add(new Connect(sc.nextInt(), sc.nextInt()));
    }

    int[] count = new int[N];

    for (int i = 1; i <= N; i++) {
      for (int j = i + 1; j <= N; j++) {
        if (friend.contains(new Connect(i, j))) continue;
        if (block.contains(new Connect(i, j))) continue;
        if (check(i, j, new ArrayList<>())) {
          count[i-1]++;
          count[j-1]++;
        }
      }
    }

    String op = "";
    for (int i : count) {
      op += (i + " ");
    }

    System.out.println(op.substring(0, op.length()-1));

    sc.close();
  }

  private static boolean check(int i, int j, List<Integer> sumi) {
    if (friend.contains(new Connect(i, j))) return true;

    boolean result = false;
    for (Connect c : friend) {
      if (c.A == i) {
        if (sumi.contains(c.B)) continue;
        List<Integer> next = new ArrayList<>(sumi);
        next.add(i);
        if (check(c.B, j, next)) {
          result = true;
          break;
        }
      }
      if (c.B == i) {
        if (sumi.contains(c.A)) continue;
        List<Integer> next = new ArrayList<>(sumi);
        next.add(i);
        if (check(c.A, j, next)) {
          result = true;
          break;
        }
      }
    }

    return result;
  }
}


class Connect {
  public int A;
  public int B;

  public Connect(int a, int b) {
    A = a;
    B = b;
  }

  @Override
  public boolean equals(Object x) {
    if (!(x instanceof Connect)) return false;
    Connect c = (Connect)x;

    if (this.A == c.A && this.B == c.B) {
      return true;
    }

    if (this.B == c.A && this.A == c.B) {
      return true;
    }

    return false;
  }
}