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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>

typedef struct {
  long head;
  long prev;
} S_DATA;

int main(void) {

  long n,m,k;
  scanf("%ld %ld %ld", &n, &m, &k);
  long **friend,**block;
  friend = (long **)malloc(sizeof (long*) * (n+1));
  block = (long **)malloc(sizeof (long*) * (n+1));
  for (long i = 0; i < n+1; i++) {
    friend[i] = (long *)malloc(sizeof (long) * (n+1));
    friend[i][0] = 0;
    block[i] = (long *)malloc(sizeof (long) * (n+1));
    block[i][0] = 0;
  }
  for (long i = 1; i <= n; i++) {
    friend[0][i] = 0;
    block[0][i] = 0;
  }
  long a,b;
  for (long i = 0; i < m; i++) {
    scanf("%ld %ld", &a, &b);
    friend[a][0]++;
    friend[a][friend[a][0]] = b;
    friend[b][0]++;
    friend[b][friend[b][0]] = a;
  }
  for (long i = 0; i < k; i++) {
    scanf("%ld %ld", &a, &b);
    block[a][0]++;
    block[a][block[a][0]] = b;
    block[b][0]++;
    block[b][block[b][0]] = a;
  }
  S_DATA users[n+1];
  long group_size[n+1];
  for (long i = 0; i <= n; i++) {
    users[i].head = 0;
    users[i].prev = 0;
    group_size[i] = 0;
  }
  users[1].head = 1;
  group_size[0] = 1;
  group_size[1] = 1;
  long current = 1;
  while (1) {
    friend[0][current]++;
    if (friend[0][current] <= friend[current][0]) {
      if (users[friend[current][friend[0][current]]].head == 0) {
        users[friend[current][friend[0][current]]].head = users[current].head;
        group_size[users[current].head]++;
        group_size[0]++;
        if (group_size[0] == n) {
          break;
        }
        users[friend[current][friend[0][current]]].prev = current;
        current = friend[current][friend[0][current]];
      } else {
        continue;
      }
    } else {
      if (users[current].prev == 0) {
        for (long i = users[current].head+1; i <= n; i++) {
          if (users[i].head == 0) {
            users[i].head = i;
            current = i;
            group_size[current]++;
            group_size[0]++;
            break;
          }
        }
      } else {
        current = users[current].prev;
      }
    }
  }
  long suggestion;
  for (long i = 1; i <= n; i++) {
    suggestion = group_size[users[i].head]-friend[i][0]-1;
    for (long j = 1; j <= block[i][0]; j++) {
      if (users[block[i][j]].head == users[i].head) {
        suggestion--;
      }
    }
    if (i != n) {
      printf("%ld ", suggestion);
    } else {
      printf("%ld\n", suggestion);
    }
  }

  return 0;
}