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.io.PrintWriter;
import java.net.ConnectException;
import java.rmi.dgc.Lease;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.NoSuchElementException;
import java.util.Objects;
import java.util.Set;

public class Main {
    public static void main(String[] args) {
        Main main = new Main();
        main.solve();
        main.out.close();
    }
    // ======================================================================
    UnionFind UF;
    Map<Integer, Set<Integer>> F = new HashMap<>(); 
    Map<Integer, Set<Integer>> B = new HashMap<>(); 
    public void solve() {
    	int N = ni();
    	int M = ni();
    	int K = ni();
    	UF = new UnionFind(N);
    	int a, b;
    	Set<Integer> set;
    	for (int i = 0; i < M; i++) {			// 友達の設定
			a = ni() - 1;
			b = ni() - 1;
			UF.connect(a, b);
			set = F.get(a);
			if(set == null) {
				set = new HashSet<Integer>();
				F.put(a, set);
			}
			set.add(b);
			set = F.get(b);
			if(set == null) {
				set = new HashSet<Integer>();
				F.put(b, set);
			}
			set.add(a);
		}
    	for (int i = 0; i < K; i++) {		// ブロックの設定
			a = ni() - 1;
			b = ni() - 1;
			set = B.get(a);
			if(set == null) {
				set = new HashSet<Integer>();
				B.put(a, set);
			}
			set.add(b);
			set = B.get(b);
			if(set == null) {
				set = new HashSet<Integer>();
				B.put(b, set);
			}
			set.add(a);
		}
    	List<Integer> list;
    	Set<Integer> setF, setB;
    	int ans;
    	for (int i = 0; i < N; i++) {
    		list = UF.member(i);					// 同じ島
    		if(list == null || list.isEmpty()) {
    			out.print(0 + " ");
    			continue;
    		}
    		ans = 0;
			for(int mem : list) {
				setF = F.get(i);
				setB = B.get(i);
				if(i != mem && (setF == null || !setF.contains(mem))
							&& (setB == null || !B.get(i).contains(mem))) {
					ans++;
				}
			}
			out.print(ans + " ");
		}
    	out.println("");
    }
	class UnionFind {
		int[] par;	    			// [子番号] = 親番号(親の場合、メンバ数のマイナス値)
									// [1]=3 [2]=-1 [3]=-2  なら　１　と　３　は　同じグループ
		public UnionFind(int n) {
			par = new int[n];
			Arrays.fill(par, -1);	// 全部が親で、メンバ数を　１　とする
		}
		// 親の番号を返す（この中で親に直接つなぎ変えている）
		int root(int x) {			
			if(par[x] < 0)     	return x;	// 親の場合はマイナスが入っている
			else                return par[x] = root(par[x]);
		}
		// 同じグループなら true を返す
		boolean same(int x, int y) {	
			return root(x) == root(y);
		}
		// グループ数を返す
		int groupCnt() {
			int ans = 0;
			for(int i = 0; i < par.length; i++) {
				if(par[i] < 0)	ans++;
			}
			return ans;
		}
		// メンバ数を返す
		int size(int x) {
			return -par[root(x)];
		}
		// メンバのリストを返す
		List<Integer> member(int x) {
			List<Integer> ans = new ArrayList<>();
			int root = root(x);
			for (int i = 0; i < par.length; i++) {
				if(root(i) == root) {
					ans.add(i);
				}
			}
			return ans;
		}
		// 同じ親につなげる
		void connect(int x, int y) {	
			x = root(x);			// 親を取得する
			y = root(y);			// 親を取得する
			if(x == y)  return;		// 同じなので処理しない
			if(size(x) > size(y)) {	// メンバ数が多い方にくっつける
				par[x] += par[y];	//   メンバ数を更新する
				par[y] = x;			//   親番号を設定する
			} else {
				par[y] += par[x];
				par[x] = y;
			}
		}
		void dump(PrintWriter out) {
			for(int i = 0; i < par.length; i++) {
			    out.println("par[" + i + "] = " + par[i]);
			}
			out.println("");
		}
	}
    // ------------------------------------------
    // ライブラリ
    // ------------------------------------------
    // Print
	private PrintWriter out = new PrintWriter(System.out);

    // Scanner
    private FastScanner scan = new FastScanner();
    int ni(){ return scan.nextInt();}
    int[] ni(int n){int[] a = new int[n]; for(int i = 0; i < n; i++){a[i] = ni();} return a;}
    int[][] ni(int y, int x){int[][] a = new int[y][x];
        for(int i = 0; i < y; i++){for(int j = 0; j < x; j++){a[i][j] = ni();}} return a;}
    long nl(){return scan.nextLong();}
    long[] nl(int n){long[] a = new long[n]; for(int i = 0; i < n; i++){a[i] = nl();} return a;}
    long[][] nl(int y, int x){long[][] a = new long[y][x];
        for(int i = 0; i < y; i++){for(int j = 0; j < x; j++){a[i][j] = nl();}} return a;}
    String ns(){return scan.next();}
    String[] ns(int n){String[] a = new String[n]; for(int i = 0; i < n; i++){a[i] = ns();} return a;}
    String[][] ns(int y, int x){String[][] a = new String[y][x];
        for(int i = 0; i < y; i++){for(int j = 0; j < x; j++){a[i][j] = ns();}} return a;}
}
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 || nl > Integer.MAX_VALUE) throw new NumberFormatException();
        return (int) nl;
    }
    public double nextDouble() { return Double.parseDouble(next());}
}