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.IOException;
import java.io.InputStream;
import java.util.*;

public class Main {
	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		int n = sc.nextInt();
		int m = sc.nextInt();
		int k = sc.nextInt();
		int[] a = new int[m];
		int[] b = new int[m];
		int[] c = new int[k];
		int[] d = new int[k];
		for (int i = 0; i < m; i++) {
			a[i] = sc.nextInt() - 1;
			b[i] = sc.nextInt() - 1;
		}
		for (int i = 0; i < k; i++) {
			c[i] = sc.nextInt() - 1;
			d[i] = sc.nextInt() - 1;
		}
		UnionFind uf = new UnionFind(n);
		for (int i = 0; i < m; i++) {
			uf.unite(a[i], b[i]);
		}
		TreeMap<Integer, TreeSet<Integer>> friends = new TreeMap<>();
		for (int i = 0; i < n; i++) {
			friends.put(i, new TreeSet<>());
		}
		for (int i = 0; i < m; i++) {
			friends.get(a[i]).add(b[i]);
			friends.get(b[i]).add(a[i]);
		}
		TreeMap<Integer, TreeSet<Integer>> blocks = new TreeMap<>();
		for (int i = 0; i < n; i++) {
			blocks.put(i, new TreeSet<>());
		}
		for (int i = 0; i < k; i++) {
			blocks.get(c[i]).add(d[i]);
			blocks.get(d[i]).add(c[i]);
		}
		int[] ans = new int[n];
		for (int i = 0; i < n; i++) {
			for (int j = i + 1; j < n; j++) {
				if (!friends.get(i).contains(j) && !blocks.get(i).contains(j) && uf.same(i, j)) {
					ans[i]++;
					ans[j]++;
				}
			}
		}
		for (int i = 0; i < n; i++) {
			System.out.print(ans[i]);
			if (i == n - 1) {
				System.out.println();
			} else {
				System.out.print(" ");
			}
		}
	}
}

class UnionFind {
	int[] par;
	int[] rank;
	int[] size;

	UnionFind(int n) {
		par = new int[n];
		rank = new int[n];
		size = new int[n];
		for (int i = 0; i < n; i++) {
			par[i] = i;
			rank[i] = 0;
			size[i] = 1;
		}
	}

	int find(int x) {
		if (par[x] == x) {
			return x;
		}
		return par[x] = find(par[x]);
	}

	void unite(int x, int y) {
		x = find(x);
		y = find(y);
		if (x == y) {
			return;
		}
		if (rank[x] < rank[y]) {
			par[x] = y;
			size[y] += size[x];
		} else {
			par[y] = x;
			size[x] += size[y];
			if (rank[x] == rank[y]) {
				rank[x]++;
			}
		}
	}

	boolean same(int x, int y) {
		return find(x) == find(y);
	}

	int size(int x) {
		return size[find(x)];
	}
}

class FastScanner {
	private final InputStream in = System.in;
	private final byte[] buffer = new byte[1024];
	private int ptr = 0;
	private int buflen = 0;
	private boolean hasNextByte() {
		if (ptr < buflen) {
			return true;
		} else {
			ptr = 0;
			try {
				buflen = in.read(buffer);
			} catch (IOException e) {
				e.printStackTrace();
			}
			if (buflen <= 0) {
				return false;
			}
		}
		return true;
	}
	private int readByte() {
		if (hasNextByte()) {
			return buffer[ptr++];
		} else {
			return -1;
		}
	}
	private static boolean isPrintableChar(int c) {
		return 33 <= c && c <= 126;
	}
	public boolean hasNext() {
		while (hasNextByte() && !isPrintableChar(buffer[ptr])) {
			ptr++;
		}
		return hasNextByte();
	}
	public String next() {
		if (!hasNext()) {
			throw new NoSuchElementException();
		}
		StringBuilder sb = new StringBuilder();
		int b = readByte();
		while (isPrintableChar(b)) {
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}
	public long nextLong() {
		if (!hasNext()) {
			throw new NoSuchElementException();
		}
		long n = 0;
		boolean minus = false;
		int b = readByte();
		if (b == '-') {
			minus = true;
			b = readByte();
		}
		if (b < '0' || '9' < b) {
			throw new NumberFormatException();
		}
		while (true) {
			if ('0' <= b && b <= '9') {
				n *= 10;
				n += b - '0';
			} else if (b == -1 || !isPrintableChar(b)) {
				return minus ? -n : n;
			} else {
				throw new NumberFormatException();
			}
			b = readByte();
		}
	}
	public int nextInt() {
		long nl = nextLong();
		if (nl < Integer.MIN_VALUE || Integer.MAX_VALUE < nl) {
			throw new NumberFormatException();
		}
		return (int) nl;
	}
	public double nextDouble() {
		return Double.parseDouble(next());
	}
}