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;
using System.Linq;
using System.Text;

public class MainClass
{
	public const long Giri = 1000000007;
	public const long DpInf = 99999999999999;
	
	public class Person
	{
		public int Index {get;set;}
		public HashSet<int> Friends {get;set;}
		
	}
	public static void Main(string[] args)
	{
		var p = Input().ToInt();
		var n = p[0];
		var m = p[1];
		var k = p[2];
		var people = new Person[n];
		
		var friendTree = new UnionFind(n);
		
		for(int i = 0; i < n; i++)
		{
			people[i] = new Person(){Index=i, Friends=new HashSet<int>()};
		}
		
		for(int i = 0; i < m; i++)
		{
			p = Input().ToInt(-1);
			people[p[0]].Friends.Add(p[1]);
			people[p[1]].Friends.Add(p[0]);
			friendTree.Unite(p[0], p[1]);
		}
		
		
		for(int i = 0; i < k; i++)
		{
			p = Input().ToInt(-1);
			if(!friendTree.InSameTree(p[0], p[1]))
				continue;
			if(!people[p[0]].Friends.Contains(p[1]))
				people[p[0]].Friends.Add(p[1]);
			if(!people[p[1]].Friends.Contains(p[0]))
				people[p[1]].Friends.Add(p[0]);
		}
		var map = new Dictionary<int, int>();
		for(int i = 0; i < n; i++)
		{
			int key = friendTree.Root(i);
			if(!map.ContainsKey(key))
				map.Add(key, 0);
			map[key]++;
		}
		
		for(int i = 0; i < n; i++)
		{
			var pe = people[i];
			Console.Write("{0} ", map[friendTree.Root(i)] - 1 - pe.Friends.Count);
		}
		Console.WriteLine();
		
	}
	
	static long GCD(long a, long b)
	{
		var v = new[] { a, b };
		while (v[1] != 0) { v = new[] { v[1], v[0] % v[1] }; }
		return v[0];
	}
	public static string[] Input() { return Console.ReadLine().Split(); }
}

public static class Smarter
{
	public static int ToInt(this string str) { return int.Parse(str); }
	public static long ToLong(this string str) { return long.Parse(str); }
	public static int[] ToInt(this string[] str, int offset = 0)
	{
		var r = new int[str.Length];
		for(int i = 0; i < str.Length; i++) r[i] = str[i].ToInt() + offset;
		return r;
	}
}

public class UnionFind
{
	private int[] par;
	public UnionFind(int n)
	{
		par = new int[n];
		for(int i = 0; i < n; i++)
			par[i] = i;
	}
	public int Root(int x)
	{
		if(par[x] == x) return x;
		return par[x] = Root(par[x]);
	}
	public void Unite(int x, int y)
	{
		int rx = Root(x);
		int ry = Root(y);
		if(rx == ry) return;
		par[rx] = ry;
	}
	public bool InSameTree(int x, int y)
	{
		return Root(x) == Root(y);
	}
}