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: Runtime Error)

import java.util.ArrayList;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Scanner;

public class Main {
	public static void main(String[] args){
		try(Scanner sc = new Scanner(System.in)) {

			int n = sc.nextInt();
			int m = sc.nextInt(); //friends
			int k = sc.nextInt(); //block

			Map<Integer, List<Integer>> friendsMap = new HashMap<>();
			Map<Integer, List<Integer>> blockMap = new HashMap<>();

			for( int i = 0 ; i < m ; i++ ) {
				int a = sc.nextInt();
				int b = sc.nextInt();
				putRelation(friendsMap, a, b);
				putRelation(friendsMap, b, a);
			}

			for(int i = 0 ; i < k ; i++ ) {
				int c = sc.nextInt();
				int d = sc.nextInt();
				putRelation(blockMap, c, d);
				putRelation(blockMap, d, c);
			}

			//
			for(int i = 0 ; i < n ; i++ ) {
				//find friends Cluster
				// initial
				HashSet<Integer> currentCluster = new HashSet<>(friendsMap.get(i+1));
				HashSet<Integer> candidates = new HashSet<>();
				
				
				// add myself
				currentCluster.add(i+1);
				// merge
				for(;;) {
					List<Integer> nextCluster = new ArrayList<Integer>();
					for(Integer f : currentCluster) {
						for(Integer ff : friendsMap.get(f)) {
							if ( !currentCluster.contains(ff) ) {
								candidates.add(ff);
								nextCluster.add(ff);
							}
						}
					}
					currentCluster.addAll(nextCluster);
					
					if ( nextCluster.size() == 0 ) {
						break;
					}
				}
				
				List<Integer> blocks = blockMap.get(i+1);
				int count = 0;
				for(Integer f : candidates) {
					if (blocks != null &&  blocks.contains(f) ) {
					}
					else {
						count++;
					}
				}
				System.out.println(count);
			}
			

		}
	}

	private static void putRelation(Map<Integer, List<Integer>> map, int key, int value) {
		if ( !map.containsKey(key) ) {
			map.put(key, new ArrayList<Integer>());
		}
		map.get(key).add(value);

	}
}