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;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using static System.Math;
using static System.Console;


class Program
{

    public class UnionFind
    {
        public UnionFind(int number)
        {
            Data = new List<int>();
            for (int i = 0; i < number; i++) Data.Add(-1);
        }

        /// <summary>
        /// 負なら親(絶対値はサイズ)　正なら子(絶対値は親のID)
        /// </summary>
        public List<int> Data { get; }

        public int Find(int node)
        {
            //自身が親なら、そのまま返す
            if (Data[node] < 0) return node;

            //子なら、その親を呼び、さらにその親を探す。再帰的に呼び出しrootを見つけ、全ての子の親をrootにする
            return Data[node] = Find(Data[node]);
        }

        public bool Unite(int node1,int node2)
        {
            //二つの点の親をさがす
            node1 = Find(node1);
            node2 = Find(node2);

            //二つの親が一致しているなら、することはない
            if(node1==node2)return false;

            int parent = 0;
            int newSon = 0;

            //サイズが小さいほうが、大きい方の子になる(親のDataの値は負であることに注意)
            if (Data[node1] > Data[node2])
            {
                parent = node2;
                newSon = node1;
            }
            else
            {
                parent = node1;
                newSon = node2;
            }

            Data[parent] += Data[newSon];
            Data[newSon] = parent;

            return true;
        }
    }

    static void Main()
    {
        int[] NMK = ReadIntN();
        int N = NMK[0];
        int M = NMK[1];
        int K = NMK[2];

        UnionFind uf = new UnionFind(N);

        List<int>[] friends = new List<int>[N];
        List<int>[] blocks = new List<int>[N];//同じUnionの中での相互ブロックのみ
        for (int i = 0; i < blocks.Length; i++)
        {
            blocks[i] = new List<int>();
            friends[i] = new List<int>();
        }


        for (int i = 0; i < M; i++)
        {
            int[] ab = ReadIntN();
            uf.Unite(ab[0]-1, ab[1]-1);
            friends[ab[0]-1].Add(ab[1]-1);
            friends[ab[1]-1].Add(ab[0]-1);
        }

        for (int i = 0; i < K; i++)
        {
            int[] cd = ReadIntN();

            if (uf.Find(cd[0]-1) == uf.Find(cd[1]-1))
            {
                blocks[cd[0]-1].Add(cd[1]-1);
                blocks[cd[1]-1].Add(cd[0]-1);
            }
        }

        int[] ans = new int[N];

        for (int i = 0; i < N; i++)
        {
            int population = uf.Data[uf.Find(i)]*-1;
            ans[i] = population - friends[i].Count - blocks[i].Count -1;
        }

        WriteLine(string.Join(" ",ans));
        ReadLine();
    }





    #region Reader
    static string ReadStr()
    {
        return Console.ReadLine();
    }
    static string[] ReadStrN()
    {
        return Console.ReadLine().Split(' ');
    }
    static int ReadInt()
    {
        return Convert.ToInt32(Console.ReadLine());
    }
    static int[] ReadIntN()
    {
        return Console.ReadLine().Split(' ').Select(s => Convert.ToInt32(s)).ToArray();
    }
    static long ReadLong()
    {
        return Convert.ToInt64(Console.ReadLine());
    }
    static long[] ReadLongN()
    {
        return Console.ReadLine().Split(' ').Select(s => Convert.ToInt64(s)).ToArray();
    }
    static long[,] ReadLongMatrix(int r, int c)
    {
        var mat = new long[r, c];
        for (int i = 0; i < r; ++i)
        {
            var x = ReadLongN();
            for (int j = 0; j < c; ++j)
            {
                mat[i, j] = x[j];
            }
        }
        return mat;
    }
    static Tuple<string, string>[] ReadTupleStr(int n)
    {
        List<Tuple<string, string>> list = new List<Tuple<string, string>>();
        for (int i = 0; i < n; ++i)
        {
            var ab = ReadStrN();
            list.Add(new Tuple<string, string>(ab[0], ab[1]));
        }
        return list.ToArray();
    }
    static Tuple<long, long>[] ReadTupleLong(int n)
    {
        List<Tuple<long, long>> list = new List<Tuple<long, long>>();
        for (int i = 0; i < n; ++i)
        {
            var ab = ReadLongN();
            list.Add(new Tuple<long, long>(ab[0], ab[1]));
        }
        return list.ToArray();
    }
    static Tuple<int, int>[] ReadTupleInt(int n)
    {
        List<Tuple<int, int>> list = new List<Tuple<int, int>>();
        for (int i = 0; i < n; ++i)
        {
            var ab = ReadIntN();
            list.Add(new Tuple<int, int>(ab[0], ab[1]));
        }
        return list.ToArray();
    }
    static Tuple<double, double>[] ReadTupleDouble(int n)
    {
        List<Tuple<double, double>> list = new List<Tuple<double, double>>();
        for (int i = 0; i < n; ++i)
        {
            var ab = ReadLongN();
            list.Add(new Tuple<double, double>(ab[0], ab[1]));
        }
        return list.ToArray();
    }
    static Dictionary<int, HashSet<int>> ReadAdjacencyList(int N, int M)
    {
        Dictionary<int, HashSet<int>> dict = new Dictionary<int, HashSet<int>>();
        for (int i = 0; i <= N; ++i)
        {
            dict.Add(i, new HashSet<int>());
        }
        for (int i = 0; i < M; ++i)
        {
            var ab = ReadIntN();
            dict[ab[0]].Add(ab[1]);
            dict[ab[1]].Add(ab[0]);
        }
        return dict;
    }
    #endregion
}