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: Runtime Error)

#include<bits/stdc++.h>
using namespace std;
struct UnionFind {
  vector<int> parent;
  vector<int> per_size;
  
  UnionFind(int n): parent(n), per_size(n) {
    for(int i = 0; i < n; ++i){
      parent[i] = i;
      per_size[i] = 1;
    }
  }

  int root(int x){
    if(parent[x] == x) return x;
    else parent[x] = root(parent[x]);
  }

  void unite(int x, int y){
    int rx = root(x);
    int ry = root(y);
    if(rx == ry) return;
    if(per_size[rx] < per_size[ry]) swap(rx, ry);
    per_size[rx] += per_size[ry];
    parent[ry] = rx;
  }

  bool is_same(int x, int y){
    return root(x) == root(y);
  }

  int get_size(int x){
    return per_size[root(x)];
  }

};

int main(){
  int n, m, k;
  cin >> n >> m >> k;
  UnionFind relation(n);
  int friends[100001];
  vector<int> block[100001];
  //vector<int> friends(n, 0);
  //vector<int> block[n];
  for(int i = 0, x, y; i < m; ++i){
    cin >> x >> y;
    x--;
    y--;
    friends[x]++;
    friends[y]++;
    relation.unite(x, y);
  }
  for(int i = 0, x, y; i < k; ++i){
    cin >> x >> y;
    x--;
    y--;
    block[x].push_back(y);
    block[y].push_back(x);
  }
  int ans;
  for(int i = 0; i < n; ++i){
    ans = relation.get_size(i) - friends[i] - 1;
    for(auto tgt: block[i]){
      if(relation.is_same(tgt,i)) ans--;
    }
    cout << (i ? " " : "") << ans;
  }
  cout << endl;
}