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
import scala.collection.mutable.ListBuffer

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 = (for {
      _ <- 0 until m
    } yield {
      in.readLine().split("\\s").map(_.toInt)
    }).map {
      case Array(a, b) =>
        if (a > b) {
          Array(b, a)
        } else {
          Array(a, b)
        }
    }.sortWith {
      case (Array(a1, b1), Array(a2, b2)) =>
        if (a1 < a2) {
          true
        } else if (a1 > a2) {
          false
        } else {
          b1 < b2
        }
    }

    val blocks = (for {
      _ <- 0 until k
    } yield {
      in.readLine().split("\\s").map(_.toInt)
    }).map {
      case Array(c, d) =>
        if (c > d) {
          Array(d, c)
        } else {
          Array(c, d)
        }
    }.sortWith {
      case (Array(c1, d1), Array(c2, d2)) =>
        if (c1 < c2) {
          true
        } else if (c1 > c2) {
          false
        } else {
          d1 < d2
        }
    }

    val person2Cluster = mutable.HashMap.empty[Int, ListBuffer[Int]]
    val person2Friends = mutable.HashMap.empty[Int, Array[Int]]
    val clusters = ListBuffer.empty[ListBuffer[Int]]

    for {
      i <- 1 to n
      if !person2Cluster.contains(i)
    } yield {
      val cluster = ListBuffer.empty[Int]
      clusters.append(cluster)
      cluster.append(i)
      person2Cluster.update(i, cluster)

      val edges = new mutable.Stack[Int]
      edges.push(i)

      while (edges.nonEmpty) {
        val p = edges.pop()
        val fs = (
          friends.filter(_ (0) == p).map(_ (1)) ++ friends.filter(_ (1) == p).map(_ (0))
          ).toArray
        person2Friends.update(p, fs)

        fs.filter { f =>
          !person2Cluster.contains(f)
        }.foreach { f =>
          cluster.append(f)
          edges.push(f)
          person2Cluster.update(f, cluster)
        }
      }
    }

    out.println(
      (for {
        p <- 1 to n
      } yield {
        val cluster = person2Cluster(p)
        val friends = person2Friends(p)
        val bs = (blocks.filter(_ (0) == p).map(_ (1)) ++ blocks.filter(_ (1) == p).map(_ (0))).filter { b =>
          cluster.contains(b)
        }
        cluster.size - friends.length - bs.size - 1
      }).mkString(" ")
    )
    out.flush()
  }
}