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;
typedef long long ll;
int g[100010];
vector<int> to[100010];

class UnionFind {
  public:
    // 親の番号を格納する。親だった場合は-(その集合のサイズ)
    vector<int> Parent;

    // 作る時はParentの値を全て-1にする
    // こうすると全てバラバラが表現できる。
    UnionFind(int N) {
      Parent = vector<int>(N, -1);
    }

    // Aがどのグループに属しているか調べる
    int root(int A) {
      if(Parent[A] < 0) return A;
      return Parent[A] = root(Parent[A]);
    }

    int size(int A) {
      return -Parent[root(A)]; // 親を取ってきたい
    }

    bool connect(int A, int B) {
      // AとBを直接繋ぐのではなく、root(A)をroot(B)をくっつける
      A = root(A);
      B = root(B);
      if(A == B) {
        //すでにくっついているからくっつけない
        return false;
      }

      // 大きい方に小さい方をくっつけたい
      // 代償が逆だったらひっくり返す
      if(size(A) < size(B)) swap(A, B);
      // Aのsizeを更新
      Parent[A] += Parent[B];
      // Bの親をAに変更する
      Parent[B] = A;

      return true;
    }

    bool same(int A, int B) {
      return root(A) == root(B);
    }
};

int main() {
  int N, M, K; cin >> N >> M >> K;
  UnionFind uni(N);
  for(int i = 0; i < M; i++) {
    int a, b; cin >> a >> b; a--, b--;
    g[a]++; g[b]++;
    uni.connect(a, b);
  }

  for(int i = 0; i < K; i++) {
    int a, b; cin >> a >> b; a--, b--;
    to[a].push_back(b);
    to[b].push_back(a);
  }

  for(int i = 0; i < N; i++) {
    int ans = uni.size(i) - 1 - g[i];
    for(int u: to[i]) {
      if(uni.same(i, u)) ans--;
    }
    cout << ans;
    if(i == N-1) cout << endl;
    else cout << " ";
  }
  return 0;
}