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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

import java.util.*

fun main(args: Array<String>) {
    val sc = Scanner(System.`in`)
    val n = sc.nextInt()
    val m = sc.nextInt()
    val k = sc.nextInt()
    val ab = (0 until m).map { sc.next().toInt() to sc.next().toInt() }
    val cd = (0 until k).map { sc.next().toInt() to sc.next().toInt() }
    println(problem157d(n, m, k, ab, cd))
}

fun problem157d(
    n: Int,
    m: Int,
    k: Int,
    ab: List<Pair<Int, Int>>,
    cd: List<Pair<Int, Int>>
): String {
    class UnionFind(n: Int) {
        val par = (0..n).toMutableList()
        val size = IntArray(n + 1) { 1 }

        fun root(x: Int): Int {
            if (par[x] == x) return x
            return root(par[x])
        }

        fun merge(x: Int, y: Int) {
            val rx = root(x)
            val ry = root(y)
            if (rx == ry) return
            par[rx] = ry
            size[rx] =+ size[ry]
        }

        fun isSame(x: Int, y: Int): Boolean {
            return root(x) == root(y)
        }

        fun size(x: Int) = par[root(x)]

        override fun toString(): String {
            return par.toString()
        }
    }

    val dame = Array<MutableSet<Int>>(n + 1) { mutableSetOf() }
    val uf = UnionFind(n)
    for (i in 0 until m) {
        val (a, b) = ab[i]
        dame[a].add(b)
        dame[b].add(a)
        uf.merge(a, b)
    }
    for (i in 0 until k) {
        val (c, d) = cd[i]
        if (!uf.isSame(c, d)) continue
        dame[c].add(d)
        dame[d].add(c)
    }
    val ans = mutableListOf<Int>()
    for (i in 1..n) {
        val mem = uf.size(i) - 1
        ans.add(mem - dame[i].size)
    }
    return ans.joinToString(" ")
}