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 (Clang 3.8.0) (judge verdict: Accepted)

#include <iostream>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <queue>
#include <stack>
#include <vector>
#include <algorithm>
#include <map>

int parent[100005];
int length[100005];
void init(){
  for (int i=0; i<100005; i++){
    parent[i] = -1;
    length[i] = 1;
  }
  return;
}
int getroot(int a){
  int temp = a;
  while(parent[temp]!=-1){
    temp = parent[temp];
  }
  return temp;
}

bool samegroup(int a, int b){
  int aroot = getroot(a);
  int broot = getroot(b);
  if (parent[a]!=aroot && parent[a]!=-1){
    parent[a] = aroot;
  }
  if (parent[b]!=broot && parent[b]!=-1){
    parent[b] = broot;
  }
  if (aroot==broot){
    return true;
  } else {
    return false;
  }
}

void combine(int a, int b){
  if (samegroup(a, b)){
    //printf("samwgourp!\n");
    return;
  }
  //printf("combine claled\n");
  int aroot = getroot(a);
  int broot = getroot(b);
  if (parent[a]!=aroot && parent[a]!=-1){
    parent[a] = aroot;
  }
  if (parent[b]!=broot && parent[b]!=-1){
    parent[b] = broot;
  }
  //  printf("root of %d = %d\n",a,aroot);
  //printf("root of %d = %d\n",b,broot);
  if (length[aroot]>length[broot]){
    parent[broot] = aroot;
    length[aroot]+=length[broot];
  } else {
    parent[aroot] = broot;
    length[broot] += length[aroot];
    //    if (length[aroot]==length[broot]){
    //      length[broot]++;
    //    }
  }
  return;
}

int count(int a){
  int aroot = getroot(a);
  if (parent[a]!=aroot && parent[a]!=-1){
    parent[a] = aroot;
  }
  //printf("root of %d = %d\n",a,aroot);
  return length[aroot];
}


int main(){
  init();
  int N, M, K;
  scanf("%d", &N);
  scanf("%d", &M);
  scanf("%d", &K);
  //printf("nmk end\n");
  std::vector<int > A(M), B(M);
  std::vector<int > C(K), D(K);
  std::vector<int > like[100005];
  for (int i=0; i<M; i++){
    scanf("%d", &A[i]);
    scanf("%d", &B[i]);
    A[i]--;
    B[i]--;
    like[A[i]].push_back(B[i]);
    like[B[i]].push_back(A[i]);
    combine(A[i], B[i]);
  }
  //printf("m end\n");
  std::vector<int > dislike[100005];
  //printf("step1\n");
  for (int i=0; i<K; i++){
    scanf("%d", &C[i]);
    scanf("%d", &D[i]);
    //printf("i=%d\n",i);
    C[i]--;
    D[i]--;
    dislike[C[i]].push_back(D[i]);
    dislike[D[i]].push_back(C[i]);
  }
  //printf("step2\n");
  for (int i=0; i<N; i++){
    //printf("====== i=%d ====== \n",i);
    int groupnum = count(i);
    //printf("groupnum=%d\n",groupnum);
    int dislikecount = 0;
    for (int j=0; j<dislike[i].size(); j++){
      if (samegroup(i, dislike[i][j])){
        dislikecount++;
      }
    }
    //printf("dislikecount=%d\n",dislikecount);
    int likecount = 0;
    for (int j=0; j<like[i].size(); j++){
      if (samegroup(i, like[i][j])){
        likecount++;
      }
    }
    //printf("likecount=%d\n",likecount);
    printf("%d\n",groupnum-1-dislikecount-likecount);
  }
  return 0;
}