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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<vector>
#include<algorithm>

using namespace std;

class UnionFind {
private:
	//ノードの数
	int size = 0;

	//ノードの親(根は自分自身)
	vector<int> par;

	//ノードの深さ(0～size)
	vector<int> rank;

	//キー値を入れる(必須ではない)
	vector<int> key;

public:
	vector<int> num;

public:
	//サイズを指定し、UnionFind木を作成し、init() を呼ぶ
	UnionFind(int V, int K = 0);

	//UnionFind木を初期化する(全てを分解する)
	void init();

	int get_size(int x);

	//ノードが属する部分木の根を求める
	int find(int x);

	//二つの部分木を合体する
	void unite(int x, int y);

	//二つのノードが同じ部分木に属しているか調べる
	bool same(int x, int y);

	//キー値をセットする
	void set_key(int index, int x);

	//キー値を取得する
	int get_key(int index);

	//デバッグ用
	void debug();
};

UnionFind::UnionFind(int V, int K) {
	if (V < 0) { V = 0; }
	size = V;
	par = vector<int>(V);
	rank = vector<int>(V);
	key = vector<int>(V, K);
	num = vector<int>(V, 1);
	init();
}

void UnionFind::init() {
	for (int i = 0; i < size; i++) {
		par[i] = i;
		rank[i] = 0;
		key[i] = 0;
		num[i] = 1;
	}
}

int UnionFind::find(int x) {
	//cout << "x = " << x << endl;
	if (par[x] == x) {
		return x;
	}
	else {
		return par[x] = find(par[x]);
	}
}

void UnionFind::unite(int x, int y) {
	//cout << "x = " << x << " y = " << y << endl;
	x = find(x);
	y = find(y);
	if (x == y) { return; }
	int xs = num[x];
	int ys = num[y];
	num[x] += ys;
	num[y] += xs;
	if (rank[x] < rank[y]) {
		par[x] = y;
	}
	else {
		par[y] = x;
		if (rank[x] == rank[y]) { rank[x]++; }
	}
}

int UnionFind::get_size(int x) {
	int n = find(x);
	return num[n];
}

bool UnionFind::same(int x, int y) {
	return find(x) == find(y);
}

void UnionFind::set_key(int index, int x) {
	key[index] = x;
}

int UnionFind::get_key(int index) {
	return key[index];
}

void UnionFind::debug() {
	cout << "UnionFindTree:" << endl;
	cout << "size = " << size << endl;
	for (int i = 0; i < size; i++) {
		cout << i << "番ノード:" << flush;
		cout << "値:" << get_key(i) << flush;
		cout << "親:" << find(i) << endl;
	}
}

int N, M, K;

int main() {
	cin >> N >> M >> K;
	UnionFind tree(N);
	vector<int> sum(N, 0);//各ユーザーの友好関係やブロック関係の数
	for (int i = 0; i < M; i++) {
		int A, B;
		cin >> A >> B;
		A--; B--;
		sum[A]++; sum[B]++;
		tree.unite(A, B);
	}
	for (int i = 0; i < K; i++) {
		int C, D;
		cin >> C >> D;
		C--; D--;
		if (tree.same(C, D)) {
			sum[C]++;
			sum[D]++;
		}
	}
	for (int i = 0; i < N; i++) {
		//cout << i+1 << "人目" << endl;
		//cout << tree.get_size(i) << endl;
		//cout << sum[i] << endl;
		int ans = max(0, tree.get_size(i) - sum[i] - 1);
		cout << ans << endl;
	}
}