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 Scala (2.11.7) (judge verdict: Accepted)

import io.StdIn._

object Main extends App {
  val firstLine = readLine.split(" ").map(_.toInt)
  val N = firstLine(0)
  val M = firstLine(1)
  val K = firstLine(2)

  val uf = UnionFind.of(N)
  val friendCount = Array.fill(N)(0)
  val blocks = Array.fill(N)(Array.empty[Int])

  for (_ <- 0 until M) {
    val friendship = readLine.split(" ").map(_.toInt - 1) // 入力は 1始まり、プログラム上は0始まり
    val a = friendship(0);
    val b = friendship(1);
    uf.unite(a, b)
    friendCount(a) = friendCount(a) + 1
    friendCount(b) = friendCount(b) + 1
  }
  for (_ <- 0 until K) {
    val blockship = readLine.split(" ").map(_.toInt - 1)
    val a = blockship(0)
    val b = blockship(1)
    blocks(a) = blocks(a) :+ b
    blocks(b) = blocks(b) :+ a
  }

  for (i <- 0 until N) {
    val blockCount = blocks(i).count(block => uf.same(i, block))
    val ans = uf.size(i) - 1 - friendCount(i) - blockCount
    print(s"${ans} ")
  }
  print("\n")
}

/**
 * d(x) == a として
 * x がツリーの根であれば: a はツリーのサイズ（ノード数) に -1 をかけた数
 * x がツリーの根でなければ: a は関連ノードのインデックス番号
 *
 * @param d
 */
case class UnionFind(d: Array[Int]) {
  def root(x: Int): Int = if (d(x) < 0) x else {
    val rt = root(d(x))
    d(x) = rt // memoization
    rt
  }

  def unite(x: Int, y: Int): Boolean = {
    val xRoot = root(x)
    val yRoot = root(y)

    if (xRoot == yRoot) false
    else {
      if (xRoot < yRoot) {
        d(xRoot) = d(xRoot) + d(yRoot)
        d(yRoot) = xRoot
      } else {
        d(yRoot) = d(yRoot) + d(xRoot)
        d(xRoot) = yRoot
      }
      true
    }
  }

  def same(x: Int, y: Int): Boolean = root(x) == root(y)

  def size(x: Int): Int = -1 * d(root(x))
}

object UnionFind {
  def of(n: Int) = UnionFind(Array.fill[Int](n)(-1))
}