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;

class Program
{
    static void Main(string[] args)
    {
        int[] arrM = Console.ReadLine().Split(' ').Select(v => int.Parse(v)).ToArray();
        int N = arrM[0];
        int M = arrM[1];
        int K = arrM[2];
        var uf = new UnionFind(N);
        var frend = new List<int>[N];
        var block = new List<int>[N];
        for (int i = 0; i < N; i++)
        {
            frend[i] = new List<int>();
            block[i] = new List<int>();
        }
        // 友達関係
        for (int i = 0; i < M; i++)
        {
            arrM = Console.ReadLine().Split(' ').Select(v => int.Parse(v)).ToArray();
            var a = arrM[0] - 1;
            var b = arrM[1] - 1;
            frend[a].Add(b);
            frend[b].Add(a);
            uf.Union(a, b);
        }
        // ブロック関係
        for (int i = 0; i < K; i++)
        {
            arrM = Console.ReadLine().Split(' ').Select(v => int.Parse(v)).ToArray();
            var c = arrM[0] - 1;
            var d = arrM[1] - 1;
            block[c].Add(d);
            block[d].Add(c);
        }
        for (int i = 0; i < N; i++)
        {
            if (i > 0) Console.Write(' ');

            // 同一グループのサイズから友達関係の数と自分自身の1を引く
            var ans = uf.Size(i) - frend[i].Count - 1;
            // 同一グループ内のブロック関係の数を引く
            foreach (var b in block[i])
            {
                if (uf.Same(i, b)) ans--;
            }
            Console.Write(ans);
        }
        Console.WriteLine();
    }

   

}

class UnionFind
{
    // 親要素のインデックスを保持する
    // 親要素が存在しない(自身がルートである)とき、マイナスでグループの要素数を持つ
    public int[] Parents { get; set; }
    public UnionFind(int n)
    {
        this.Parents = new int[n];
        for (int i = 0; i < n; i++)
        {
            // 初期状態ではそれぞれが別のグループ(ルートは自分自身)
            // ルートなのでマイナスで要素数(1個)を保持する
            this.Parents[i] = -1;
        }
    }

    // 要素xのルート要素はどれか
    public int Find(int x)
    {
        // 親がマイナスの場合は自分自身がルート
        if (this.Parents[x] < 0) return x;
        // ルートが見つかるまで再帰的に探す
        // 見つかったルートにつなぎかえる
        this.Parents[x] = Find(this.Parents[x]);
        return this.Parents[x];
    }

    // 要素xの属するグループの要素数を取得する
    public int Size(int x)
    {
        // ルート要素を取得して、サイズを取得して返す
        return -this.Parents[this.Find(x)];
    }

    // 要素x, yが同じグループかどうか判定する
    public bool Same(int x, int y)
    {
        return this.Find(x) == this.Find(y);
    }

    // 要素x, yが属するグループを同じグループにまとめる
    public bool Union(int x, int y)
    {
        // x, y のルート
        x = this.Find(x);
        y = this.Find(y);
        // すでに同じグループの場合処理しない
        if (x == y) return false;

        // 要素数が少ないグループを多いほうに書き換えたい
        if (this.Size(x) < this.Size(y))
        {
            var tmp = x;
            x = y;
            y = tmp;
        }
        // まとめる先のグループの要素数を更新
        this.Parents[x] += this.Parents[y];
        // まとめられるグループのルートの親を書き換え
        this.Parents[y] = x;
        return true;
    }
}