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.Collections.Generic;
using System.IO;
using System.Linq;
using System.Text;
using System.Text.RegularExpressions;
using static System.Math;

namespace CSharpSpiral {
	class Program {
		static void Main(string[] args) {
			var sw = new StreamWriter(Console.OpenStandardOutput()) {
				AutoFlush=
#if DEBUG
			true
#else
			false
#endif
			};
			Console.SetOut(sw);
			new Program().Solve(new ConsoleInput(
#if DEBUG
			new StreamReader("../../../inputData.txt")
#else
			Console.In
#endif
			, ' '));
			Console.Out.Flush();
		}
		public void Solve(ConsoleInput cin) {
			var N = cin.ReadInt;
			var Friendship = new UnionFindForest(N);
			var Already = new int[N];
			var M = cin.ReadInt;
			var K = cin.ReadInt;
			for(int i = 0; i<M; i++) {
				var AB = cin.ReadIntArray(2);
				AB[0]--;AB[1]--;
				Friendship.Union(AB[0], AB[1]);
				Already[AB[0]]++;
				Already[AB[1]]++;
			}
			var Blocks = new HashSet<int>[N];
			for(int i = 0; i<N; i++) 
				Blocks[i]=new HashSet<int>();
			for(int i = 0; i<K; i++) {
				var CD = cin.ReadIntArray(2);
				CD[0]--;CD[1]--;
				Blocks[CD[0]].Add(CD[1]);
				Blocks[CD[1]].Add(CD[0]);
			}
			var Cloud = new Dictionary<UnionFindForest.Node, int>();
			var Reference = new UnionFindForest.Node[N];
			for(int i = 0; i<N; i++) {
				Reference[i]=Friendship.Root(i);
				if(Cloud.ContainsKey(Reference[i])) {
					Cloud[Reference[i]]++;
				} else {
					Cloud.Add(Reference[i], 1);
				}
			}
			var Excluded = new int[N];
			for(int i = 0; i<N; i++) 
				Excluded[i]=Blocks[i].Count(j => Friendship.Find(i, j));
			var Res = new int[N];
			for(int i = 0; i<N; i++) {
				Res[i]=Cloud[Reference[i]]-Excluded[i]-1-Already[i];
			}
			Console.WriteLine(Res.Join());
		}
	}
	static class ExtendClass {
		public static void Swap<T>(this T[] A, int x, int y) {
			var tmp = A[x];
			A[x]=A[y];
			A[y]=tmp;
		}
		public static string Join<T>(this IEnumerable<T> X, string separator = " ") => string.Join(separator, X);
		public static IEnumerable<int> Indices<T>(this T[] X) => Enumerable.Range(0, X.Length);
		public static IEnumerable<int> Indices<T>(this T[,] X, int dim)
			=> Enumerable.Range(0, X.GetLength(dim));
		public static string Join<T>(this T[,] X, string separatorD2 = ",", string separatorD1 = "\n")
			=> X.Indices(0).Select(
				i => X.Indices(1).Select(j => X[i, j]).Join(separatorD2)
			).Join(separatorD1);
	}
	class ConsoleInput {
		private readonly TextReader _stream;
		private readonly char _separator = ' ';
		private readonly Queue<string> inputStream;
		public ConsoleInput(TextReader stream, char separator = ' ') {
			this._separator=separator;
			this._stream=stream;
			inputStream=new Queue<string>();
		}
		public string Read {
			get {
				if(inputStream.Count!=0) return inputStream.Dequeue();
				string[] tmp = _stream.ReadLine().Split(_separator);
				for(int i = 0; i<tmp.Length; ++i)
					inputStream.Enqueue(tmp[i]);
				return inputStream.Dequeue();
			}
		}
		public string ReadLine { get { return _stream.ReadLine(); } }
		public int ReadInt { get { return int.Parse(Read); } }
		public long ReadLong { get { return long.Parse(Read); } }
		public double ReadDouble { get { return double.Parse(Read); } }
		public string[] ReadStrArray(long N) {
			var ret = new string[N];
			for(long i = 0; i<N; ++i)
				ret[i]=Read;
			return ret;
		}
		public int[] ReadIntArray(long N) { var ret = new int[N]; for(long i = 0; i<N; ++i) ret[i]=ReadInt; return ret; }
		public long[] ReadLongArray(long N) { var ret = new long[N]; for(long i = 0; i<N; ++i) ret[i]=ReadLong; return ret; }
		public double[] ReadDoubleArray(long N) { var ret = new double[N]; for(long i = 0; i<N; ++i) ret[i]=ReadDouble; return ret; }
		public bool isEOF {
			get { return _stream.Peek()==-1; }
		}
	}
	class UnionFindForest {
		public class Node {
			public int value { get; set; }
			public Node parent { get; set; }
			public Node root {
				get {
					var path = new Queue<Node>();
					var res = this;
					while(res.parent!=res) {
						path.Enqueue(res);
						res=res.parent;
					}
					while(path.Any())
						path.Dequeue().parent=res;
					return res;
				}
			}
			public int rank { get; private set; }    //meanful iff this is root
			public void rankIncrement() => rank++;
			public Node(int thisValue) {
				value=thisValue;
				parent=this;
				rank=1;
			}
		}

		private readonly Dictionary<int, Node> phoneBook = new Dictionary<int, Node>();
		public void Union(int x, int y) => Union(phoneBook[x], phoneBook[y]);
		public void Union(Node X, Node Y) {
			var RX = X.root;
			var RY = Y.root;
			switch(RX.rank.CompareTo(RY.rank)) {
				case 1:
					RY.parent=RX;
					break;
				case -1:
					RX.parent=RY;
					break;
				default:
					RX.parent=RY;
					RY.rankIncrement();
					break;
			}
		}
		public bool Find(int x, int y) => Find(phoneBook[x], phoneBook[y]);
		public bool Find(Node X, Node Y) => X.root==Y.root;
		public void Add(int id) {
			var N = new Node(id);
			phoneBook.Add(id, N);
		}
		public Node Root(int id) => phoneBook[id].root;
		public UnionFindForest(int N) {
			for(int i = 0; i<N; i++) 
				Add(i);
		}
	}
}