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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Numerics;
using System.Linq;
using System.Collections.Generic;
class ProgramD
    {
        static void Main(string[] args)
        {
            
            //入力
            string[] input = Console.ReadLine().Split(' ');
            int n = int.Parse(input[0]);
            int m = int.Parse(input[1]);
            int k = int.Parse(input[2]);
 
            int[] count = new int[n];
 
            UnionFind tree = new UnionFind();
 
            tree.par = new int[n];
            tree.temp = new int[n];
            tree.reset(n);
 
            for(int i = 0;i < m;i++)
            {
                string[] input2 = Console.ReadLine().Split(' ');
                int x = int.Parse(input2[0]);
                int y = int.Parse(input2[1]);
                tree.unite(x-1, y-1); // xの木とyの木を併合する
                count[x-1]--;
                count[y-1]--;
            }
        
            for(int i= 0;i< n;i++)
              count[i] += tree.size(i) - 1;
 
            for(int i = 0;i < k;i++)
            {
                string[] input3 = Console.ReadLine().Split(' ');
                int x = int.Parse(input3[0]);
                int y = int.Parse(input3[1]);
                if(tree.same(x-1,y-1) == true)
                {
                    count[x-1]--;
                    count[y-1]--;
                }
            }
 
            
 
            for(int i = 0;i < n;i++)
            {
                Console.Write(count[i]);
                Console.Write(" ");
            }
        }
    }
 
    class UnionFind
    {
 
        public int[] par;
        public int[] temp;
 
        public void reset(int n)
        {
            for(int i = 0;i < n;i++)
                par[i] = -1;
        }
        public int root(int x)
        {
            if(par[x] < 0)
                return x;
            else
                return root(par[x]);
        }
 
        public void unite(int x, int y)
        { // xとyの木を併合
            int rx = root(x); //xの根をrx
            int 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
            if(par[ry] > par[rx])
            {
                 int temp = ry;
                 ry = rx;
                 rx = temp;
            }
            par[ry] += par[rx];
            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
        }
 
        public bool same(int x, int y) 
        { // 2つのデータx, yが属する木が同じならtrueを返す
            int rx = root(x);
            int ry = root(y);
            return rx == ry;
        }
        
        public int size(int x)
        {
            return -par[root(x)];
        }
      
        public int[] member(int n) 
        {
            for(int i = 0;i < n;i++)
            {
              int r = root(i);
              temp[r]++;
            }
            return temp;
        }
    }