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 <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <string>
#include <sstream>
#include <complex>
#include <vector>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <set>
#include <fstream>
#include <numeric>
using namespace std;
typedef long long int ll;

#define EPS (1e-7)
#define INF 1e18
#define max(p,q)((p)>(q)?(p):(q))
#define min(p,q)((p)<(q)?(p):(q))
#define PI (acos(-1))

#define REP(i, n) for(int i = 0; i < (int)(n); i++)
#define rep(i, init, n) for(int i = init; i <(int)(n); i++)

struct UnionFind {
    vector<ll> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2
    vector<ll> _size; // 属する連結成分の大きさ

    UnionFind(ll N) : par(N), _size(N) { //最初は全てが根であるとして初期化
        for(ll i = 0; i < N; i++) {
            par[i] = i;
            _size[i] = 1;
        }
    }

    ll root(ll x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }

    void unite(ll x, ll y) { // xとyの木を併合
        ll rx = root(x); //xの根をrx
        ll ry = root(y); //yの根をry
        if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
        par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
        _size[ry] += _size[rx];
    }

    bool same(ll x, ll y) { // 2つのデータx, yが属する木が同じならtrueを返す
        ll rx = root(x);
        ll ry = root(y);
        return rx == ry;
    }

    ll size(ll x){
        ll rx = root(x);
        return _size[rx];
    }
};

int main() {

    ll N, M, K;
    cin >> N >> M >> K;
    vector<ll> f(N, 0), b(N, 0);
    UnionFind U = UnionFind(N);
    ll k, l;
    REP(i, M){
        cin >> k >> l;
        k--;
        l--;
        f[k]++;
        f[l]++;
        U.unite(k, l);
    }
    REP(i, K){
        cin >> k >> l;
        k--;
        l--;
        if (U.same(k, l)){
            b[k]++;
            b[l]++;
        }
        
    }
    REP(i, N){
        ll tmp = U.size(i) - f[i] - b[i] - 1;
        cout << tmp << " ";
    }
    cout << endl;

    return 0;
}