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 9.2.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
int *fG[100000];
int *bG[100000];
int label[100000], mcnts[100000] = {0}, mpos[100000] = {0}, kcnts[100000] = {0},
				   kpos[100000] = {0}, a[100000] = {0}, b[100000] = {0},
				   c[100000] = {0}, d[100000] = {0};
int nextL = 1;
void dfs(int node, int l) {
	if (label[node] != 0)
		return;
	int i = 0;
	label[node] = l;
	while (fG[node][i] != -1) {
		dfs(fG[node][i], l);
		i++;
	}
	if (nextL == l)
		nextL++;
	return;
}
int main() {
	int n, m, k;
	scanf("%d%d%d", &n, &m, &k);
	for (int i = 0; i < m; i++) {
		scanf("%d%d", &a[i], &b[i]);
		a[i]--;
		b[i]--;
		mcnts[a[i]]++;
		mcnts[b[i]]++;
	}
	for (int i = 0; i < n; i++) {
		fG[i] = malloc(sizeof(int) * (mcnts[i] + 1));
		if (fG[i] == NULL) {
			return -1;
		}
	}
	for (int i = 0; i < m; i++) {
		fG[a[i]][mpos[a[i]]] = b[i];
		mpos[a[i]]++;
		fG[b[i]][mpos[b[i]]] = a[i];
		mpos[b[i]]++;
	}
	for (int i = 0; i < n; i++) {
		fG[i][mpos[i]] = -1;
	}
	for (int i = 0; i < k; i++) {
		scanf("%d%d", &c[i], &d[i]);
		c[i]--;
		d[i]--;
		kcnts[c[i]]++;
		kcnts[d[i]]++;
	}
	for (int i = 0; i < n; i++) {
		bG[i] = malloc(sizeof(int) * (kcnts[i] + 1));
		if (bG[i] == NULL) {
			return -1;
		}
	}
	for (int i = 0; i < k; i++) {
		bG[c[i]][kpos[c[i]]] = d[i];
		kpos[c[i]]++;
		bG[d[i]][kpos[d[i]]] = c[i];
		kpos[d[i]]++;
	}
	for (int i = 0; i < n; i++) {
		bG[i][kpos[i]] = -1;
	}

	for (int i = 0; i < n; i++) {
		dfs(i, nextL);
	}
	int labelCnt[100000] = {0};
	for (int i = 0; i < n; i++) {
		// printf("%d ", label[i]);
		labelCnt[label[i] - 1]++;
	}
	// printf("\n");
	for (int i = 0; i < n; i++) {
		int ans, fCnt = 0, fbCnt = 0, bCnt = 0;
		while (fG[i][fCnt] != -1)
			fCnt++;
		while (bG[i][bCnt] != -1) {
			if (label[bG[i][bCnt]] == label[i])
				fbCnt++;
			bCnt++;
		}
		// printf("l:%d f:%d b:%d ", labelCnt[label[i] - 1], fCnt, fbCnt);
		ans = labelCnt[label[i] - 1] - fCnt - fbCnt - 1;
		printf("%d ", ans);
	}
	return 0;
}