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 Java (OpenJDK 11.0.6) (judge verdict: Time Limit Exceeded)

import java.util.*;

public class Main {
    static int max = 100000;
    static int[] stack = new int[max];
    static int top = 0;
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        int k = sc.nextInt();
        TreeMap<Integer,TreeSet<Integer>[]> relation = new TreeMap<>();
        for (int i=1;i<=n;i++){
            relation.put(i,new TreeSet[2]);
            relation.get(i)[0] = new TreeSet<>();
            relation.get(i)[1] = new TreeSet<>();
        }
        for (int i=0;i<m;i++){
            int a = sc.nextInt();
            int b = sc.nextInt();
            relation.get(a)[0].add(b);
            relation.get(b)[0].add(a);
        }
        for (int i=0;i<k;i++){
            int c = sc.nextInt();
            int d = sc.nextInt();
            relation.get(c)[1].add(d);
            relation.get(d)[1].add(c);
        }
        HashSet<HashSet<Integer>> nest = new HashSet<>();
        for (int i=1;i<=n;i++){
            init();
            boolean b = true;
            HashSet<Integer> team = new HashSet<>();
            for (HashSet<Integer> h : nest){
                if (h.contains(i)){
                    team.addAll(h);
                    b = false;
                }
            }
            if (b){
                HashSet<Integer> hash = new HashSet<>();
                hash.add(i);
                boolean[] seen = new boolean[n+1];
                for (int j=1;j<=n;j++)seen[i]=false;
                push(i);
                seen[i]=true;
                while (!isEmpty()){
                    int v = pop();
                    for (int l : relation.get(v)[0]){
                        if (!seen[l]){
                            seen[l]=true;
                            push(l);
                            hash.add(l);
                        }
                    }
                }
                nest.add(hash);
                team.addAll(hash);
            }
            team.removeAll(relation.get(i)[0]);
            team.removeAll(relation.get(i)[1]);
            System.out.print((team.size()-1)+" ");
        }
        System.out.println();
    }
    static void init(){top=0;}
    static boolean isEmpty(){return (top==0);}
    static boolean isFull(){return (top==max);}

    static void push(int v){
        if (isFull()){
            System.out.println("error: stack is full.");
            return;
        }
        stack[top++] = v;
    }
    static int pop(){
        if (isEmpty()){
            System.out.println("error: stack is empty");
            return -1;
        }
        return stack[--top];
    }
}