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.io.InputStream;
import java.io.PrintStream;
import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;

public class Main {
    public static void main(String[] args) throws Exception {
        solve(System.in, System.out);
    }

    static class Root {
        int root;
        int size;
        Root (int root) {
            this.root = root;
            this.size = 1;
        }
    }

    static Root find(int i) {
        if(roots[i].root == i) {
            return roots[i];
        }
        roots[i].root = roots[roots[i].root].root;
        return find(roots[i].root);
    }

    static void union(Root a, Root b) {
        if(a.root == b.root) return;
        a.root = b.root;
        b.size += a.size;
    }

    static Root[] roots;

    static void solve(InputStream is, PrintStream os) {
        // Your code here!
        Scanner scan = new Scanner(is);
        int N = scan.nextInt();
        int M = scan.nextInt();
        int K = scan.nextInt();
        roots = new Root[N];
        int[] friends = new int[N];
        int[] blocks = new int[N];
        for(int i = 0; i < N; i++) {
            roots[i] = new Root(i);
        }

        for(int i = 0; i < M; i++) {
            int A = scan.nextInt()-1;
            int B = scan.nextInt()-1;
            friends[A]++;
            friends[B]++;
            Root rootA = find(A);
            Root rootB = find(B);
            union(rootA, rootB);
        }

        for(int i = 0; i < K; i++) {
            int C = scan.nextInt()-1;
            int D = scan.nextInt()-1;
            Root rootC = find(roots[C].root);
            Root rootD = find(roots[D].root);
            if(rootC.root == rootD.root) {
                blocks[C]++;
                blocks[D]++;
            }
        }

        for(int i = 0; i < N; i++) {
            Root root = find(i);
            int possible = root.size - friends[i] - blocks[i] - 1;
            os.print(possible);
            if(i != N-1)
                os.print(" ");
        }
        os.println();
    }
}