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: Accepted)


import java.util.ArrayList;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Main main = new Main();
		main.run();
	}

	class UF{
		int p[];
		int rank[];
		int cnt[];
		public UF(int n) {
			p = new int[n];
			rank = new int[n];
			cnt = new int[n];

			for(int i=0; i<n; i++) {
				p[i] = i;
				rank[i]=1;
				cnt[i]=1;
			}
		}

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

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

		public void unite(int x, int y) {
			link(find(x), find(y));
		}

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

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

	public void run() {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		int k = sc.nextInt();

		UF uf = new UF(n);
		int f[] = new int[n];
		for(int i=0; i<m; i++) {
			int a = sc.nextInt(); a--;
			int b = sc.nextInt(); b--;
			uf.unite(a,b);
			f[a]++;
			f[b]++;
		}
		ArrayList<ArrayList<Integer>> blk = new ArrayList<ArrayList<Integer>>();
		for(int i=0; i<n; i++) {
			blk.add(new ArrayList<Integer>());
		}
		for(int i=0; i<k; i++) {
			int c = sc.nextInt(); c--;
			int d = sc.nextInt(); d--;
			blk.get(c).add(d);
			blk.get(d).add(c);
		}

		for(int i=0; i<n; i++) {
			int ans = uf.size(i) - f[i] - 1;
			for(int b : blk.get(i)) {
				if(uf.same(i, b)) {
					ans--;
				}
			}
			System.out.print(ans + (i==n-1 ? "\n" : " "));
		}
		sc.close();
	}

}