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.*;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int numberOfUsers = scanner.nextInt();
        int friendships = scanner.nextInt();
        int blockships = scanner.nextInt();

        int[][] relationshipPlan = new int[numberOfUsers][numberOfUsers];
        int[] answer = new int[numberOfUsers];
        for (int i = 0; i < friendships; i++) {
            int friendshipOne = scanner.nextInt();
            int friendshipTwo = scanner.nextInt();
            relationshipPlan[friendshipOne - 1][friendshipTwo - 1] = 1;
            relationshipPlan[friendshipTwo - 1][friendshipOne - 1] = 1;
        }

        for (int i = 0; i < blockships; i++) {
            int blockshipOne = scanner.nextInt();
            int blockshipTwo = scanner.nextInt();
            relationshipPlan[blockshipOne - 1][blockshipTwo - 1] = -1;
            relationshipPlan[blockshipTwo - 1][blockshipOne - 1] = -1;
        }

        List<List<Integer>> relationships = new ArrayList<>();
        List<List<Integer>> blockedUsers = new ArrayList<>();

        for (int i = 0; i < numberOfUsers; i++) {
            List<Integer> friends = new ArrayList<>();
            List<Integer> blocked = new ArrayList<>();

            for (int k = 0; k < numberOfUsers; k++) {
                if (relationshipPlan[i][k] == 1) {
                    friends.add(k + 1);
                } else if (relationshipPlan[i][k] == -1) {
                    blocked.add(k + 1);
                }
            }
            relationships.add(friends);
            blockedUsers.add(blocked);
        }


        Set<Integer> candidates = new HashSet<>();
        Set<Integer> visited = new HashSet<>();
        List<Integer> relationship;
        for (int i = 0; i < relationships.size(); i++) {
            candidates.clear();
            visited.clear();

            relationship = relationships.get(i);
            lookForFriends(relationship, relationships, candidates, visited);
            int tempAnswer = 0;
            for (int candidate : candidates) {
                if (!relationship.contains(candidate) && !blockedUsers.get(i).contains(candidate) && candidate != i + 1) {
                    tempAnswer++;
                }
            }
            answer[i] = tempAnswer;
        }


        StringBuilder stringBuilder = new StringBuilder();
        for (int i = 0; i < numberOfUsers; i++) {
            stringBuilder.append(answer[i] + " ");
        }

        System.out.println(stringBuilder.toString().trim());
    }

    private static void lookForFriends(List<Integer> currentUser, List<List<Integer>> allUsers, Set<Integer> candidates, Set<Integer> visited) {
        for (int friend : currentUser) {
            candidates.add(friend);
            if (!visited.contains(friend)) {
                visited.add(friend);
                lookForFriends(allUsers.get(friend - 1), allUsers, candidates, visited);
            }
        }
    }
}