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.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        int k = sc.nextInt();
        HashSet<Integer>[] friends = new HashSet[n];
        HashSet<Integer>[] blocks = new HashSet[n];
        for (int i = 0; i < n; i++) {
            friends[i] = new HashSet<>();
            blocks[i] = new HashSet<>();
        }
        UnionFindTree uft = new UnionFindTree(n);
        for (int i = 0; i < m; i++) {
            int a = sc.nextInt() - 1;
            int b = sc.nextInt() - 1;
            friends[Math.min(a, b)].add(Math.max(a, b));
            uft.unite(a, b);
        }
        for (int i = 0; i < k; i++) {
            int a = sc.nextInt() - 1;
            int b = sc.nextInt() - 1;
            blocks[Math.min(a, b)].add(Math.max(a, b));
        }
        int[] counts = new int[n];
        HashMap<Integer, ArrayList<Integer>> map = uft.getSet();
        for (ArrayList<Integer> list : map.values()) {
            for (int i = 0; i < list.size() - 1; i++) {
                for (int j = i + 1; j < list.size(); j++) {
                    if (!friends[list.get(i)].contains(list.get(j)) && !blocks[list.get(i)].contains(list.get(j))) {
                        counts[list.get(i)]++;
                        counts[list.get(j)]++;
                    }
                }
            }
        }
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < n; i++) {
            if (i != 0) {
                sb.append(" ");
            }
            sb.append(counts[i]);
        }
		System.out.println(sb);
   }
    
    static class UnionFindTree {
        int[] parents;
        
        public UnionFindTree(int size) {
            parents = new int[size];
            for (int i = 0; i < size; i++) {
                parents[i] = i;
            }
        }
        
        public int find(int x) {
            if (parents[x] == x) {
                return x;
            } else {
                return parents[x] = find(parents[x]);
            }
        }
        
        public void unite(int x, int y) {
            int xx = find(x);
            int yy = find(y);
            if (xx == yy) {
                return;
            }
            parents[xx] = yy;
        }
        
        public HashMap<Integer, ArrayList<Integer>> getSet() {
            HashMap<Integer, ArrayList<Integer>> ans = new HashMap<>();
            for (int i = 0; i < parents.length; i++) {
                int x = find(i);
                if (ans.containsKey(x)) {
                    ans.get(x).add(i);
                } else {
                    ArrayList<Integer> list = new ArrayList<>();
                    list.add(i);
                    ans.put(x, list);
                }
            }
            return ans;
        }
    }

}