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.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Scanner;

public class Main {

	static List <ArrayList<Node>> friends;
	static int []ans ;
	
	
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		int n = Integer.parseInt(sc.next());
		int m = Integer.parseInt(sc.next());
		int k = Integer.parseInt(sc.next());
		
		Unionfind uf = new Unionfind(n);
	
		friends = new ArrayList<ArrayList<Node>>();
		
		for(int i = 0 ;i < n ;i++) {
			friends.add(new ArrayList<Node>());
		}
		
		for(int i = 0 ;i < m ;i++) {
			int a = Integer.parseInt(sc.next())-1;
			int b = Integer.parseInt(sc.next())-1;
			friends.get(a).add(new Node(b,false,true));
			friends.get(b).add(new Node(a,false,true));
			uf.union(a, b);
		}
		
		for(int i = 0 ;i < k ;i++) {
			int c = Integer.parseInt(sc.next())-1;
			int d = Integer.parseInt(sc.next())-1;
			friends.get(c).add(new Node(d,false,true));
			friends.get(d).add(new Node(c,false,true));
		}
		ans = new int [n];
		
		for(int i = 0 ; i < n ;i++) {
			for(Node e : friends.get(i)) {
				if(e.getLike()||(e.getDislike()&&uf.same(i,e.getTo()))) {
					ans[i]++;
				}
			}
		}
		
		int []size = new int [n];
		for(int i = 0 ;i < n ;i++) {
			size[i] = uf.size(i);
		}
		
		for(int i = 0 ;i < n ;i++) {
			int num = size[i] - ans[i] - 1;
			System.out.println(num);
		}
		}
	
}
class Node{
	private int to;
	private boolean like;
	private boolean dislike;
	
	public Node(int to, boolean like,boolean dislike) {
		this.to = to;
		this.like = like;
		this.dislike = dislike;
	}
	
	public int getTo() {
		return to;
	}
	public boolean getLike() {
		return like;
	}
	public boolean getDislike() {
		return dislike;
	}
}

class Unionfind {

	private int[] data;
	private int[] rootsize;
	private int size;
	
	public Unionfind(int size) {
		this.size = size;
		this.data = new int [size];
		clear();
	}
	
	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];
			data[y] = x;
		}
	}
	
	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);
	}
	
	public int size(int x) {
		int p = root(x);
		int rootsize = 0;
		
		for(int i = 0 ;i < size ;i++) {
			if(root(i) == p) {
				rootsize++;
			}			
		}
		return rootsize ;
	}
}