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;

public class Solution
{
    static bool[] visited;
    static List<int> GetFriends(HashSet<int>[] graph, int start)
    {
        var friends = new List<int>() { start };
        var currents = new List<int>() { start };
        visited[start] = true;

        while (currents.Count > 0)
        {
            var nexts = new List<int>();
            foreach (var cur in currents)
            {
                foreach (var nei in graph[cur])
                {
                    if (visited[nei])
                    {
                        continue;
                    }
                    nexts.Add(nei);
                    friends.Add(nei);
                    visited[nei] = true;
                }
            }
            currents = nexts;
        }

        return friends;
    }

    public static void Main()
    {
        var vals = Console.ReadLine().Split(' ').Select(int.Parse).ToArray();
        var N = vals[0];
        var M = vals[1];
        var K = vals[2];
        var friends = Enumerable.Range(0, N + 1).Select(_ => new HashSet<int>()).ToArray();
        var blocks = Enumerable.Range(0, N + 1).Select(_ => new HashSet<int>()).ToArray();
        visited = new bool[N + 1];

        for (int i = 0; i < M; i++)
        {
            vals = Console.ReadLine().Split(' ').Select(int.Parse).ToArray();
            var a = vals[0];
            var b = vals[1];
            friends[a].Add(b);
            friends[b].Add(a);
        }
        for (int i = 0; i < K; i++)
        {
            vals = Console.ReadLine().Split(' ').Select(int.Parse).ToArray();
            var c = vals[0];
            var d = vals[1];
            blocks[c].Add(d);
            blocks[d].Add(c);
        }

        var groups = new HashSet<int>[N + 1];
        for (int i = 1; i <= N; i++)
        {
            if (groups[i] == null)
            {
                var g = new HashSet<int>(GetFriends(friends, i));
                foreach (var f in g)
                {
                    groups[f] = g;
                }
            }
        }

        var candConts = new int[N + 1];
        for (int i = 1; i <= N; i++)
        {
            var count = groups[i].Count();
            count -= friends[i].Count(f => groups[i].Contains(f));
            count -= blocks[i].Count(f => groups[i].Contains(f));
            count--;
            candConts[i] = count;
        }

        Console.WriteLine(string.Join(" ", candConts.Skip(1)));
    }
}