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: Time Limit Exceeded)

import java.io.{BufferedOutputStream, BufferedReader, InputStreamReader, PrintWriter}

import scala.collection.mutable

object Main {
  def main(args: Array[String]): Unit = {
    val in = new BufferedReader(new InputStreamReader(System.in))
    val out = new PrintWriter(new BufferedOutputStream(System.out))
    val Array(n, m, k) = in.readLine().split("\\s").map(_.toInt)

    val friends = Array.ofDim[Array[Int]](m)
    for {
      i <- 0 until m
    } yield {
      friends(i) = in.readLine().split("\\s").map(_.toInt)
    }

    val blocks = Array.ofDim[Array[Int]](k)
    for {
      i <- 0 until k
    } yield {
      blocks(i) = in.readLine().split("\\s").map(_.toInt)
    }

    type Person = Int
    type ClusterId = Int

    val person2ClusterId = Array.ofDim[Int](n + 1)
    val person2FriendsSize = Array.ofDim[Int](n + 1)
    val clusterId2size = mutable.HashMap.empty[ClusterId, Int]

    @inline
    def myRelations(sets: Array[Array[Person]], me: Person): Array[Person] = {
      sets.filter(_ (0) == me).map(_ (1)) ++ sets.filter(_ (1) == me).map(_ (0))
    }

    @inline
    def myFriends(me: Person): Array[Person] = {
      myRelations(friends, me)
    }

    @inline
    def myBlocks(me: Person): Array[Person] = {
      myRelations(blocks, me)
    }

    var clusterNum = 0

    for {
      p <- 1 to n
      if person2ClusterId(p) == 0
    } yield {
      clusterNum += 1
      val clusterId = clusterNum
      person2ClusterId(p) = clusterId

      val edges = new mutable.Stack[Person]
      edges.push(p)
      var clusterSize = 1

      while (edges.nonEmpty) {
        val p = edges.pop()
        val fs = myFriends(p)
        person2FriendsSize(p) = fs.length

        fs.filter { f =>
          person2ClusterId(f) == 0
        }.foreach { f =>
          clusterSize += 1
          edges.push(f)
          person2ClusterId(f) = clusterId
        }
      }

      clusterId2size.update(clusterId, clusterSize)
    }

    for {
      p <- 1 to n
    } {
      val clusterId = person2ClusterId(p)
      val friendsSize = person2FriendsSize(p)
      val bpsSize = myBlocks(p).count { bp =>
        person2ClusterId(bp) == clusterId
      }
      if (p != 1) {
        out.print(" ")
      }
      out.print(clusterId2size(clusterId) - friendsSize - bpsSize - 1)
    }
    out.println()
    out.flush()
  }
}