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 {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int M = sc.nextInt();
        int K = sc.nextInt();
        String res="";
        Unionfind friend = new Unionfind(N);
        int A;
        int B;
        int[] ans = new int[N];
        int[] f = new int[N];
        for(int i=0; i<M; i++){
            A=sc.nextInt()-1;
            B=sc.nextInt()-1;
            friend.union(A, B);
            f[A]++;
            f[B]++;
        }
        for(int i=0; i<N; i++){
            ans[i] = friend.getSize(i) - f[i] - 1;
        }
        int C;
        int D;
        for(int i=0; i<K; i++){
            C=sc.nextInt()-1;
            D=sc.nextInt()-1;
            if(friend.same(C, D)){
                ans[C]--;
                ans[D]--;
            }
        }
        for(int i=0; i<N; i++){
            res = res + ans[i] + " ";
        }
        
        System.out.println(res.trim());
        
        }
     









        static class Unionfind {
		 
            private int[] data;
            private int []size;
            
            public Unionfind(int size) {
                this.data = new int [size];
                this.size = new int [size];
                for(int i = 0 ;i < size ;i++) {
                    initnode(i);
                }
                clear();
            }
            
            public void initnode(int i) {
                size[i] = 1;
            }
            
            public void clear() {
                Arrays.fill(data, -1);
            }
            
            public void union(int x , int y ) {
                if((x = root(x)) != (y = root(y))) {
                    if(data[y] < data[x]) {
                        final int t = x;
                        x = y;
                        y = t;
                    }
                    data[x] += data[y];
                    size[root(x)] += size[root(y)];
                    data[y] = x;
                }
            }
            
            public int getSize(int i) {
                return size[root(i)];
            }
            
            public int root(int x) {
                if(data[x] < 0) {
                    return x;
                }
                else {
                    return data[x] = root(data[x]);
                }
            }
            
            public boolean same(int x, int y) {
                return root(x) == root(y);
            }
            
        }
    }