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

public class Main {

	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);

		final int N = scanner.nextInt();
		final int M = scanner.nextInt();
		final int K = scanner.nextInt();

		List<Integer>[] friendLists = new ArrayList[N];
		List<Integer>[] brockLists = new ArrayList[N];
		for (int i = 0; i < N; i++) {
			friendLists[i] = new ArrayList<Integer>();
			brockLists[i] = new ArrayList<Integer>();
		}

		int A, B;
		for (int i = 0; i < M; i++) {
			A = scanner.nextInt() - 1;
			B = scanner.nextInt() - 1;
			friendLists[A].add(B);
			friendLists[B].add(A);
		}
		for (int i = 0; i < K; i++) {
			A = scanner.nextInt() - 1;
			B = scanner.nextInt() - 1;
			brockLists[A].add(B);
			brockLists[B].add(A);
		}

		boolean[] visited = new boolean[N];

		Set<Integer> group;
		List<Set<Integer>> groupList = new ArrayList<Set<Integer>>();

		for (int i = 0; i < N; i++) {
			if (!visited[i]) {
				group = new HashSet<Integer>();
				DFS(friendLists, i, visited, group);
				groupList.add(group);
			}
		}

		int result;
		for (int i = 0; i < N; i++) {
			for (Set<Integer> g : groupList) {
				if (g.contains(i)) {
					result = g.size() - 1;
					for (Integer j : g) {
						if (friendLists[i].contains(j))
							result--;
						else if (brockLists[i].contains(j))
							result--;
					}
					System.out.print(result + " ");
					break;
				}
			}
		}

	}

	static void DFS(List<Integer>[] graph, int vertex, boolean[] visited, Set<Integer> reachable) {
		if (!visited[vertex]) {
			visited[vertex] = true;
			reachable.add(vertex);
			graph[vertex].forEach(neighbor -> DFS(graph, neighbor, visited, reachable));
		}
	}

}