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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashSet;

public class Main {

	public static void main(String[] args) throws IOException {
		// TODO 自動生成されたメソッド・スタブ
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

		String[] tmpArray = br.readLine().split(" ");
		int n = Integer.parseInt(tmpArray[0]);
		int m = Integer.parseInt(tmpArray[1]);
		int k = Integer.parseInt(tmpArray[2]);

		@SuppressWarnings("unchecked")
		HashSet<Integer>[] friends = new HashSet[n];
		@SuppressWarnings("unchecked")
		HashSet<Integer>[] block = new HashSet[n];

		DisjointSet unionFind = new DisjointSet(n);

		for(int i = 0; i < n; i++){
			friends[i] = new HashSet<Integer>();
			block[i] = new HashSet<Integer>();
		}

		for(int i = 0; i < m; i++){
			tmpArray = br.readLine().split(" ");
			int a = Integer.parseInt(tmpArray[0]) - 1;
			int b = Integer.parseInt(tmpArray[1]) - 1;

			friends[a].add(b);
			friends[b].add(a);

			unionFind.union(a, b);
		}

		for(int i = 0; i < k; i++){
			tmpArray = br.readLine().split(" ");
			int c = Integer.parseInt(tmpArray[0]) - 1;
			int d = Integer.parseInt(tmpArray[1]) - 1;

			block[c].add(d);
			block[d].add(c);
		}

		for(int i = 0; i < n; i++){
			int count = 0;
			for(int j = 0; j < n; j++){
				if(i == j){
					continue;
				}
//				System.out.println("i = "+i+" j = "+j);
				if(block[i].contains(j) || friends[i].contains(j)){
//					System.out.println("friend or block");
					continue;
				}

				if(unionFind.isSameSet(i, j)){
					count++;
				}
			}
			System.out.print(count);
			if(i != n - 1){
				System.out.print(" ");
			}
		}
		System.out.println();
	}

}

class DisjointSet {
	private int n;
	private int[] p;
	private int[] rank;

	public DisjointSet(int n){
		this.n = n;

		p = new int[n + 1];
		rank = new int[n + 1];

		for(int i = 1; i <= n; i++){
			makeSet(i);
		}
	}

	private void makeSet(int x){
		p[x] = x;
		rank[x] = 0;
	}

	public void union(int x, int y){
		link (findSet(x), findSet(y));
	}

	private int findSet(int x){
		if(x != p[x]){
			p[x] = findSet( p[x]);
		}
		return p[x];
	}

	public boolean isSameSet(int x, int y){
		return findSet(x) == findSet(y);
	}

	private void link(int x, int y){
		if(rank[x] > rank[y]){
			p[y] = x;
		}
		else {
			p[x] = y;
			if(rank[x] == rank[y]){
				rank[y]++;
			}
		}
	}
}