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 scala.collection.mutable.ListBuffer

object Main extends App {
  val in = new java.util.Scanner(System.in)
  val N, M, K = in.nextInt
  val uf = new UnionFind(N + 1)
  val friends = new Array[Int](N + 1)

  (1 to M).foreach {m =>
    val a, b = in.nextInt
    uf.union(a, b)
    friends(a) += 1
    friends(b) += 1
  }

  val blocks = Array.fill(N + 1)(new ListBuffer[Int])
  (1 to K).foreach {m =>
    val c, d = in.nextInt
    blocks(c) += d
    blocks(d) += c
  }

  val result = (1 to N).map {i =>
    uf.size(i) - friends(i) - blocks(i).filter(j => uf.find(i) == uf.find(j)).size - 1
  }

  println(result.mkString(" "))
}

class UnionFind(nodes: Int) {
  val parents = Array.tabulate(nodes)(identity)
  val sizes = Array.fill(nodes)(1)

  // @scala.annotation.tailrec
  /** @retrun (root, rank) */
  def findWithRank(x: Int, rank: Int = 0): (Int, Int) = {
    parents(x) match {
      case p if x == p => (p, rank)
      case p => {
        val (newRoot, newRank) = findWithRank(p, rank + 1)
        parents(x) = newRoot
        (newRoot, newRank)
      }
    }
  }

  def find(x: Int): Int = {
    parents(x) match {
      case p if x == p => p
      case p => {
        val newRoot = find(p)
        parents(x) = newRoot
        newRoot
      }
    }
  }

  def size(x: Int) = sizes(find(x))

  def union(x: Int, y: Int): Int = {
    val (xRoot, xRank) = findWithRank(x)
    val (yRoot, yRank) = findWithRank(y)
    val newRoot = if (xRank < yRank) yRoot else xRoot
    if (xRoot != yRoot) {
      parents(xRoot) = newRoot
      parents(yRoot) = newRoot
      sizes(newRoot) = sizes(xRoot) + sizes(yRoot)
    }
    parents(x) = newRoot
    parents(y) = newRoot
    newRoot
  }
}