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.*;
import java.util.Map.Entry;
 
class Main {
	 static int mod =  (int) (Math.pow(10,9)+7);
	 static int[] ans;
//	 static int mod = 998244353;
   public static void main(String[] args) {
	   
       Scanner sc = new Scanner(System.in);
       int  N = sc.nextInt();
       int  M = sc.nextInt();
       int  K = sc.nextInt();
       
       Unionfind uf = new Unionfind(N);
       
       ans=new int[N];
       
       int [][]like = new int [M][2];
       for (int i=0 ; i<M; i++) {
    	   like[i][0] = Integer.parseInt(sc.next())-1;
    	   like[i][1] = Integer.parseInt(sc.next())-1;
    	   uf.union(like[i][0], like[i][1]);
       }
       
       int [][]dislike = new int [K][2];
       for (int i=0 ; i<K; i++) {
    	   dislike[i][0] = Integer.parseInt(sc.next())-1;
    	   dislike[i][1] = Integer.parseInt(sc.next())-1;
       }
       for(int i=0;i<N;i++) {
    	   ans[i]=uf.getSize(i)-1;
       }
       
       for(int i=0;i<M;i++) {
    	   ans[like[i][0]]--;
    	   ans[like[i][1]]--;
       }
       
       for(int i=0;i<K;i++) {
    	   if(uf.same(dislike[i][0],dislike[i][1])){
        	   ans[dislike[i][0]]--;
        	   ans[dislike[i][1]]--;
    	   }
       }
		for(int i = 0;i<N;i++) {
			System.out.println(ans[i]);
		}
   }
   

}

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);
		}
		
	}