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: Wrong Answer)



import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.List;
import java.util.Scanner;
import java.util.Set;

public class Main {
  private static List<List<Integer>> friend = new ArrayList<>();
  private static List<List<Integer>> block = new ArrayList<>();
  private static boolean[] isVisited;

  public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);
    int n = scanner.nextInt();
    int m = scanner.nextInt();
    int k = scanner.nextInt();
    isVisited = new boolean[n];
    for (int i = 0; i < n; i++) {
      friend.add(new ArrayList<>());
      block.add(new ArrayList<>());
    }
    for (int i = 0; i < m; i++) {
      int a = scanner.nextInt();
      int b = scanner.nextInt();
      friend.get(a - 1)
          .add(b-1);
      friend.get(b - 1)
          .add(a-1);
    }
    for (int i = 0; i < k; i++) {
      int c = scanner.nextInt();
      int d = scanner.nextInt();
      block.get(c - 1)
          .add(d-1);
      block.get(d - 1)
          .add(c-1);
    }

    for (int i = 0; i < n; i++) {
      int count = 0;
      for (int j : friend.get(i)) {
        if(!friend.get(i).contains(j) && !block.get(i).contains(j)) {
          Arrays.fill(isVisited,false);
          if(isCandidate(i, j)) {
            count++;
          }
        }
      }
      System.out.print(count-1 + " ");
    }
  }

  private static boolean isCandidate(final int p1, final int p2) {
    for (int i : friend.get(p1)) {
      if(!isVisited[i]) {
        isVisited[i] = true;
        if (i == p2 || isCandidate(i, p2)) {
          return true;
        }
      }
    }
    return false;
  }

}