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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.*;

public class Main {
    static InputStream is;
    static PrintWriter out;
    static String INPUT = "";
    static List<Integer>[] g;
    static int[] group;
    static int[] size;
    static int id;
    static int count;

    static void solve() {
        int n = ni();
        int m = ni();
        int k = ni();

        g = new List[n];
        group = new int[n];
        size = new int[n];
        int[][] block = new int[k][2];

        for (int i = 0; i < n; i++) {
            g[i] = new ArrayList<>();
        }

        for (int i = 0; i < m; i++) {
            int a = ni() - 1;
            int b = ni() - 1;
            g[a].add(b);
            g[b].add(a);
        }

        for (int i = 0; i < k; i++) {
            block[i][0] = ni() - 1;
            block[i][1] = ni() - 1;
        }

        Arrays.fill(group, -1);
        id = 0;
        for (int i = 0; i < n; i++) {
            if (group[i] == -1) {
                count = 0;
                dfs(i);
                size[id] = count;
                id++;
            }
        }

        int[] ans = new int[n];
        for (int i = 0; i < n; i++) {
            ans[i] = size[group[i]] - 1 - g[i].size();
        }

        for (int i = 0; i < k; i++) {
            int c = block[i][0];
            int d = block[i][1];
            if (group[c] == group[d]) {
                ans[c]--;
                ans[d]--;
            }
        }

        for (int i = 0; i < n; i++) {
            out.print(ans[i]);
            out.print((i == n - 1)? "" : " ");
        }

        out.println();
    }

    public static void dfs(int v) {
        group[v] = id;
        count++;
        for (int u : g[v]) {
            if (group[u] == -1) {
                dfs(u);
            }
        }
    }

    public static void main(String[] args) throws Exception {
        long S = System.currentTimeMillis();
        is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
        out = new PrintWriter(System.out);

        solve();
        out.flush();
        long G = System.currentTimeMillis();
        tr(G-S+"ms");
    }

    private static boolean eof() {
        if(lenbuf == -1)return true;
        int lptr = ptrbuf;
        while(lptr < lenbuf)if(!isSpaceChar(inbuf[lptr++]))return false;

        try {
            is.mark(1000);
            while(true){
                int b = is.read();
                if(b == -1){
                    is.reset();
                    return true;
                }else if(!isSpaceChar(b)){
                    is.reset();
                    return false;
                }
            }
        } catch (IOException e) {
            return true;
        }
    }

    private static byte[] inbuf = new byte[1024];
    static int lenbuf = 0, ptrbuf = 0;

    private static int readByte() {
        if(lenbuf == -1)throw new InputMismatchException();
        if(ptrbuf >= lenbuf){
            ptrbuf = 0;
            try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
            if(lenbuf <= 0)return -1;
        }
        return inbuf[ptrbuf++];
    }

    private static boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }

    private static int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }

    private static double nd() { return Double.parseDouble(ns()); }
    private static char nc() { return (char)skip(); }

    private static String ns() {
        int b = skip();
        StringBuilder sb = new StringBuilder();
        while(!(isSpaceChar(b))){
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }

    private static char[] ns(int n) {
        char[] buf = new char[n];
        int b = skip(), p = 0;
        while(p < n && !(isSpaceChar(b))){
            buf[p++] = (char)b;
            b = readByte();
        }
        return n == p ? buf : Arrays.copyOf(buf, p);
    }

    private static char[][] nm(int n, int m) {
        char[][] map = new char[n][];
        for(int i = 0;i < n;i++)map[i] = ns(m);
        return map;
    }

    private static int[] na(int n) {
        int[] a = new int[n];
        for(int i = 0;i < n;i++)a[i] = ni();
        return a;
    }

    private static int ni() {
        int num = 0, b;
        boolean minus = false;
        while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
        if(b == '-'){
            minus = true;
            b = readByte();
        }

        while(true){
            if(b >= '0' && b <= '9'){
                num = num * 10 + (b - '0');
            }else{
                return minus ? -num : num;
            }
            b = readByte();
        }
    }

    private static long nl() {
        long num = 0;
        int b;
        boolean minus = false;
        while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
        if(b == '-'){
            minus = true;
            b = readByte();
        }

        while(true){
            if(b >= '0' && b <= '9'){
                num = num * 10 + (b - '0');
            }else{
                return minus ? -num : num;
            }
            b = readByte();
        }
    }

    private static void tr(Object... o) { if(INPUT.length() != 0)System.out.println(Arrays.deepToString(o)); }

    // Union-Find
    // https://algs4.cs.princeton.edu/15uf/UF.java.html
    public static class UnionFind {
        private int count = 0;
        private int[] parent, rank;

        public UnionFind(int n) {
            count = n;
            parent = new int[n];
            rank = new int[n];
            for (int i = 0; i < n; i++) {
                parent[i] = i;
            }
        }

        public int find(int p) {
            while (p != parent[p]) {
                parent[p] = parent[parent[p]];
                p = parent[p];
            }
            return p;
        }

        public void union(int p, int q) {
            int rootP = find(p);
            int rootQ = find(q);
            if (rootP ==rootQ) return;
            if (rank[rootQ] > rank[rootP]) {
                parent[rootP] = rootQ;
            } else {
                parent[rootQ] = rootP;
                if (rank[rootP] == rank[rootQ]) {
                    rank[rootP]++;
                }
            }
            count--;
        }

        public int count() {
            return count;
        }

        public void print() {
            out.println("Parent:");
            for (int i : parent) {
                out.print(i);
                out.print(" ");
            }
            out.println();
            out.println("Rank:");
            for (int i : rank) {
                out.print(i);
                out.print(" ");
            }
            out.println();
        }
    }

    // Dijkstra
    // Return min distance from start to end O(ElogV) (negative cost is prohibited)
    // edge is int[3] array {from,to,cost}
    // edges is edge list from specific node
    // all_edges is Map<from node number,edges>
    static int dijkstra(Map<Integer,List<int[]>> all_edges,int start,int end,int max_node_number) {
        int[] distance = new int[max_node_number + 1];
        Arrays.fill(distance, -1);
        PriorityQueue<int[]> p_que = new PriorityQueue<>((a,b) -> ((distance[a[0]] + a[2]) - (distance[b[0]] + b[2])));
        distance[start] = 0;
        List<int[]> edges = all_edges.get(start);
        p_que.addAll(edges);
        while (distance[end] < 0) {
            int[] nearest_edge = p_que.poll();
            assert nearest_edge != null;
            if (distance[nearest_edge[1]] < 0) {
                distance[nearest_edge[1]] = distance[nearest_edge[0]] + nearest_edge[2];
                if (all_edges.containsKey(nearest_edge[1])) {
                    edges = all_edges.get(nearest_edge[1]);
                    for (int[] edge : edges) {
                        if (distance[edge[1]] < 0) p_que.add(edge);
                    }
                }
            }
        }
        return distance[end];
    }

    // Enumerate primes in [2, n] with O(n log log n)
    public static List<Integer> sieveOfEratosthenes(int n) {
        boolean[] isPrime = new boolean[n+1];
        Arrays.fill(isPrime, true);
        for (int i = 2; i * i <= n; i++) {
            if (isPrime[i]) {
                for (int j = i + i; j <= n; j += i) {
                    isPrime[j] = false;
                }
            }
        }
        List<Integer> primes = new ArrayList<>();
        for (int i = 2; i <= n; i++) {
            if (isPrime[i]) {
                primes.add(i);
            }
        }
        return primes;
    }

    // Count divisor
    public static int countDivisor(int n) {
        int[] list = new int[n + 1];
        Arrays.fill(list, 0);
        int num = n;
        for (int i = 2; i <= n; i++) {
            while (num % i == 0) {
                list[i] = list[i] + 1;
                num /= i;
            }
        }
        int ans = 1;
        for (int i : list) {
            ans *= (i + 1);
        }
        return ans;
    }
}