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: Accepted)

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

struct node{
	int num;
	struct node *next;
};

struct person{
	int label;
	struct node *friend;
	struct node *block;
};

void reg_friend(int m, struct person *p){
	int a, b;
	struct node *tmp;
	int i;

	for(i=0; i<m; i++){
		scanf("%d %d", &a, &b);
		a--;
		b--;
		tmp = (struct node *)malloc(sizeof(struct node));
		tmp->num = b;
		tmp->next = p[a].friend;
		p[a].friend = tmp;
		tmp = (struct node *)malloc(sizeof(struct node));
		tmp->num = a;
		tmp->next = p[b].friend;
		p[b].friend = tmp;
	}
}

void reg_block(int k, struct person *p){
	int a, b;
	struct node *tmp;
	int i;

	for(i=0; i<k; i++){
		scanf("%d %d", &a, &b);
		a--;
		b--;
		tmp = (struct node *)malloc(sizeof(struct node));
		tmp->num = b;
		tmp->next = p[a].block;
		p[a].block = tmp;
		tmp = (struct node *)malloc(sizeof(struct node));
		tmp->num = a;
		tmp->next = p[b].block;
		p[b].block = tmp;
	}
}

int dfs(struct person p[], int i, int label){
	struct node *head = NULL;
	struct node *tmp, *b, *c;
	int cnt = 1;

	p[i].label = label;
	tmp = (struct node *)malloc(sizeof(struct node));
	tmp->num = i;
	tmp->next = head;
	head = tmp;
	while(head != NULL){
		b = head;
		head = head->next;
		for(c=p[b->num].friend; c!=NULL; c=c->next){
			if(p[c->num].label == -1){
				p[c->num].label = label;
				tmp = (struct node *)malloc(sizeof(struct node));
				tmp->num = c->num;
				tmp->next = head;
				head = tmp;
				cnt++;
			}
		}
	}

	return cnt;
}

int main(){
	int n, m, k;
	struct person *p;
	struct node *q;
	int i;
	int *size;
	int cnt, now_label = 0;

	scanf("%d %d %d", &n, &m, &k);

	p = (struct person *)malloc(sizeof(struct person) * n);
	size = (int *)malloc(sizeof(int) * n);
	for(i=0; i<n; i++){
		p[i].label = -1;
		p[i].friend = NULL;
		p[i].block = NULL;
	}

	reg_friend(m, p);
	reg_block(k, p);

	for(i=0; i<n; i++){
		if(i != 0){
			printf(" ");
		}

		if(p[i].label != -1){
			cnt = size[p[i].label];
			for(q=p[i].friend; q!=NULL; q=q->next){
				cnt--;
			}
			for(q=p[i].block; q!=NULL; q=q->next){
				if(p[q->num].label == p[i].label){
					cnt--;
				}
			}
			cnt--;
			printf("%d", cnt);
			continue;
		}

		size[now_label] = dfs(p, i, now_label);
		cnt = size[now_label];
		now_label++;
		for(q=p[i].friend; q!=NULL; q=q->next){
			cnt--;
		}
		for(q=p[i].block; q!=NULL; q=q->next){
			if(p[q->num].label == p[i].label){
				cnt--;
			}
		}
		cnt--;
		printf("%d", cnt);
	}

	return 0;
}