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: Accepted)


import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Optional;
import java.util.Scanner;
import java.util.Set;

public class Main {
    public static void main(final String[] args) {
        final Scanner scanner = new Scanner(System.in);
        final int n = scanner.nextInt();
        final int m = scanner.nextInt();
        final int k = scanner.nextInt();

        final Map<Integer, List<Integer>> friendMap = new HashMap<>();
        final Map<Integer, List<Integer>> blockMap = new HashMap<>();
        final FastUnionFindTree tree = new FastUnionFindTree(n);
        for (int i = 0; i < m; i++) {
            final int a = scanner.nextInt();
            final int b = scanner.nextInt();

            friendMap.computeIfAbsent(a, v -> new ArrayList<>()).add(b);
            friendMap.computeIfAbsent(b, v -> new ArrayList<>()).add(a);
            tree.unit(Math.min(a, b), Math.max(a, b));
        }

        for (int i = 0; i < k; i++) {
            final int c = scanner.nextInt();
            final int d = scanner.nextInt();

            if (tree.isSame(c, d)) {
                blockMap.computeIfAbsent(c, v -> new ArrayList<>()).add(d);
                blockMap.computeIfAbsent(d, v -> new ArrayList<>()).add(c);
            }
        }

        final Map<Integer, List<Integer>> preFriendMap = new HashMap<>();
        for (int i = 1; i <= n; i++) {
            final int root = tree.getRoot(i);
            preFriendMap.computeIfAbsent(root, v -> new ArrayList<>()).add(i);
        }

        final int[] answers = new int[n + 1];
        for (final List<Integer> value : preFriendMap.values()) {
            for (final int i : value) {
                final List<Integer> friends = Optional.ofNullable(friendMap.get(i)).orElse(Collections.emptyList());
                final List<Integer> blocks = Optional.ofNullable(blockMap.get(i)).orElse(Collections.emptyList());

                answers[i] = value.size() - friends.size() - blocks.size() - 1;
            }
        }

        for (int i = 1; i <= n; i++) {
            System.out.print(answers[i] + " ");
        }
    }

    static class FastUnionFindTree {
        private int[] nodes;
        private final List<Integer> indices = new ArrayList<>();

        FastUnionFindTree(final int numOfNodes) {
            this.nodes = init(numOfNodes);
        }

        private int[] init(final int numOfNodes) {
            final int[] array = new int[numOfNodes + 1];
            for (int i = 1; i <= numOfNodes; i++) {
                array[i] = i;
            }

            return array;
        }

        int getRoot(final int nodeNumber) {
            final int rootNode = nodes[nodeNumber];
            if (rootNode == nodeNumber) {
                for (final Integer index : indices) {
                    nodes[index] = rootNode;
                }
                indices.clear();
                return nodeNumber;
            }

            indices.add(nodeNumber);
            return getRoot(rootNode);
        }

        boolean isSame(final int nodeA, final int nodeB) {
            return getRoot(nodeA) == getRoot(nodeB);
        }

        void unit(final int nodeA, final int nodeB) {
            final int rootA = getRoot(nodeA);
            final int rootB = getRoot(nodeB);

            if (rootA == rootB) {
                return;
            }

            nodes[rootB] = rootA;
        }
    }
}