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 C# (Mono 4.6.2.0) (judge verdict: Accepted)


using System;
using System.Collections.Generic;
class Program
{
	static void Main(string[] args)
	{
		string[] str = Console.ReadLine().Split();
		int N = int.Parse(str[0]);
		int M = int.Parse(str[1]);
		int K = int.Parse(str[2]);
		var listM = new List<int>[N];
		var listK = new List<int>[N];
		for(var i=0;i<N;i++){
			listM[i] = new List<int>();
			listK[i] = new List<int>();
		}
		var uf = new UnionFind(N);
		int[]C = new int[N];
		for(var i=0;i<N;i++){
			C[i] = 1;
		}
		for(var i=0;i<M;i++){
			string[] str2 = Console.ReadLine().Split();
			int m1 = int.Parse(str2[0]);
			int m2 = int.Parse(str2[1]);
			listM[m1-1].Add(m2-1);
			listM[m2-1].Add(m1-1);
			uf.Unite(m1-1,m2-1);
			int c1 = C[m1-1];
			int c2 = C[m2-1];
			C[m1-1] = c1+c2;
			C[m2-1] = c1+c2;
		}
		var uf2 = new UnionFind(N);
		for(var i=0;i<K;i++){
			string[] str2 = Console.ReadLine().Split();
			int m1 = int.Parse(str2[0]);
			int m2 = int.Parse(str2[1]);
			listK[m1-1].Add(m2-1);
			listK[m2-1].Add(m1-1);
			uf2.Unite(m1-1,m2-1);
		}
      	int[]D = new int[N];
		for(var i=0;i<N;i++){
          Console.Error.Write(uf.Root(i)+" ");
          D[uf.Root(i)]++;
        }
		for(var i=0;i<N;i++){
			int ct = D[uf.Root(i)]-1;
			Console.Error.Write(ct+" ");
			for(var j=0;j<listM[i].Count;j++){
				if(uf.Root(i)==uf.Root(listM[i][j])){
					ct--;
				}
			}
			for(var j=0;j<listK[i].Count;j++){
				if(uf.Root(i)==uf.Root(listK[i][j])){
					ct--;
				}
			}
			Console.Write(ct+" ");
		}
	}
}
public class UnionFind
{
    private int[] data;
	public UnionFind(int size){
        data = new int[size];
        for (int i = 0; i < size; i++) data[i] = -1;
    }
    public bool Unite(int x, int y){
        x = Root(x);
        y = Root(y);
        if (x != y){
            if (data[y] < data[x]){
                var tmp = y;
                y = x;
                x = tmp;
            }
            data[x] += data[y];
            data[y] = x;
        }
        return x != y;
    }
	public bool IsSameGroup(int x, int y){
        return Root(x) == Root(y);
    }
    public int Root(int x){
        return data[x] < 0 ? x : data[x] = Root(data[x]);
    }
}