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 Go (1.6) (judge verdict: Accepted)

package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
	"strings"
)

func main() {
	io := NewIo()
	defer io.Flush()

	n, m, k := io.Int(), io.Int(), io.Int()

	u := NewUnionFind(n)

	edges := map[int][]int{}
	for i := 0; i < m; i++ {
		a, b := io.Int()-1, io.Int()-1
		edges[a] = append(edges[a], b)
		edges[b] = append(edges[b], a)
		u.Unite(a, b)
	}

	blockEdges := map[int][]int{}
	for i := 0; i < k; i++ {
		c, d := io.Int()-1, io.Int()-1
		blockEdges[c] = append(blockEdges[c], d)
		blockEdges[d] = append(blockEdges[d], c)
	}

	anss := make([]int, n)

	groups := u.Groups()
	for _, g := range groups {
		for _, idx := range g {
			anss[idx] = len(g) - len(edges[idx]) - 1
			for _, v := range blockEdges[idx] {
				if u.Same(idx, v) {
					anss[idx]--
				}
			}
		}
	}

	io.PrintIntLn(anss)

}

type Io struct {
	reader    *bufio.Reader
	writer    *bufio.Writer
	tokens    []string
	nextToken int
}

func NewIo() *Io {
	return &Io{
		reader: bufio.NewReader(os.Stdin),
		writer: bufio.NewWriter(os.Stdout),
	}
}

func (io *Io) Flush() {
	err := io.writer.Flush()
	if err != nil {
		panic(err)
	}
}

func (io *Io) NextLine() string {
	var buffer []byte
	for {
		line, isPrefix, err := io.reader.ReadLine()
		if err != nil {
			panic(err)
		}
		buffer = append(buffer, line...)
		if !isPrefix {
			break
		}
	}
	return string(buffer)
}

func (io *Io) Next() string {
	for io.nextToken >= len(io.tokens) {
		line := io.NextLine()
		io.tokens = strings.Fields(line)
		io.nextToken = 0
	}
	r := io.tokens[io.nextToken]
	io.nextToken++
	return r
}

func (io *Io) Int() int {
	i, err := strconv.Atoi(io.Next())
	if err != nil {
		panic(err)
	}
	return i
}
func (io *Io) PrintLn(a ...interface{}) {
	fmt.Fprintln(io.writer, a...)
}

func (io *Io) PrintIntLn(a []int) {
	b := []interface{}{}
	for _, x := range a {
		b = append(b, x)
	}
	io.PrintLn(b...)
}

type UnionFind struct {
	parents []int
}

func NewUnionFind(n int) *UnionFind {
	parent := make([]int, n)
	for i := 0; i < n; i++ {
		parent[i] = i
	}
	return &UnionFind{
		parents: parent,
	}
}

func (u *UnionFind) Root(x int) int {
	if u.parents[x] == x {
		return x
	}
	u.parents[x] = u.Root(u.parents[x])
	return u.parents[x]
}

func (u *UnionFind) Unite(x, y int) {
	rx := u.Root(x)
	ry := u.Root(y)
	if rx == ry {
		return
	}
	if rx < ry {
		u.parents[rx] = ry
	} else {
		u.parents[ry] = rx
	}
}

func (u *UnionFind) Same(x, y int) bool {
	rx := u.Root(x)
	ry := u.Root(y)
	if rx == ry {
		return true
	}
	return false
}

func (u *UnionFind) Size(x int) int {
	root := u.Root(x)
	size := 0
	for _, p := range u.parents {
		if u.Root(p) == root {
			size++
		}
	}
	return size
}

func (u *UnionFind) Groups() map[int][]int {
	groups := map[int][]int{}

	for i, p := range u.parents {
		groups[u.Root(p)] = append(groups[u.Root(p)], i)
	}
	return groups
}