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>
#define rep(i,l,n) for(int i=l;i<n;i++)

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

struct person{
	int group;	// 所属するグループの親番号, 親であればそのグループのサイズ*(-1)
	struct relation *friends;
	struct relation *block;
};

int insert(struct person*,int,int,int);
int unit(struct person*,int,int);
int find(struct person*,int);
int same(struct person*,int,int);

int main(){
	int N,M,K;
	scanf("%d %d %d",&N,&M,&K);
	struct person *data = (struct person*)malloc(sizeof(struct person)*N);
	rep(i,0,N){
		(data+i)->group = -1;
		(data+i)->friends = (struct relation*)malloc(sizeof(struct relation));
		((data+i)->friends)->num = i;
		((data+i)->friends)->next = NULL;
		(data+i)->block = (struct relation*)malloc(sizeof(struct relation));
		((data+i)->block)->num = i;
		((data+i)->block)->next = NULL;
	}
	rep(i,0,M){
		int a,b;
		scanf("%d %d",&a,&b);
		a--; b--;
		insert(data,a,b,0);
		if(same(data,a,b)==0) unit(data,a,b);
	}
	rep(i,0,K){
		int a,b;
		scanf("%d %d",&a,&b);
		a--; b--;
		insert(data,a,b,1);
	}
	rep(i,0,N){
		int a = find(data,i);
		int count = -(data+a)->group;
		struct relation *data1 = (data+i)->friends;
		while( data1 != NULL ){
			if( same(data,i,data1->num) ) count--;
			data1 = data1->next;
		}
		data1 = (data+i)->block;
		while( data1 != NULL ){
			if( same(data,i,data1->num) ) count--;
			data1 = data1->next;
		}
		printf("%d ",count+1);
	}
	return 0;
}
/*
int insert(struct person *data0,int a,int b,int c){
	struct relation *data1;
	if( c==0 ) data1 = (data0+a)->friends;
	else data1 = (data0+a)->block;
	while( data1->next != NULL ) data1 = data1->next;
	data1->next = (struct relation*)malloc(sizeof(struct relation));
	(data1->next)->num = b;
	(data1->next)->next = NULL;
	data1 = (data0+b)->friends;
	while( data1->next != NULL ) data1 = data1->next;
	data1->next = (struct relation*)malloc(sizeof(struct relation));
	(data1->next)->num = a;
	(data1->next)->next = NULL;
	return 0;
}*/

int insert(struct person *data0,int a,int b,int c){
	struct relation *data1;
	if( c==0 ){
		data1 = (data0+a)->friends;
		(data0+a)->friends = (struct relation*)malloc(sizeof(struct relation));
		((data0+a)->friends)->num = b;
		((data0+a)->friends)->next = data1;
		data1 = (data0+b)->friends;
		(data0+b)->friends = (struct relation*)malloc(sizeof(struct relation));
		((data0+b)->friends)->num = a;
		((data0+b)->friends)->next = data1;
	}else{
		data1 = (data0+a)->block;
		(data0+a)->block = (struct relation*)malloc(sizeof(struct relation));
		((data0+a)->block)->num = b;
		((data0+a)->block)->next = data1;
		data1 = (data0+b)->block;
		(data0+b)->block = (struct relation*)malloc(sizeof(struct relation));
		((data0+b)->block)->num = a;
		((data0+b)->block)->next = data1;
	}
	return 0;
}

int unit(struct person *data,int a,int b){
	int *data1 = &(data+a)->group;
	int *data2 = &(data+b)->group;
	if( *data1 >= 0 ){	// data+a が子
		int c = find(data,a);
		return unit(data,c,b);
	}
	if( *data2 >= 0 ){	// data+b が子
		int c = find(data,b);
		return unit(data,a,c);
	}
	// 両方親
	if( *data1 < *data2 ){	// aの方がサイズが大きい
		*data1 += *data2;
		*data2 = a;
		return 0;
	}
	*data2 += *data1;
	*data1 = b;
	return 0;
}

int find(struct person *data,int a){	// data+aの親番号を返却
	if( (data+a)->group < 0 ) return a;
	(data+a)->group = find(data,(data+a)->group);
	return (data+a)->group;
}

int same(struct person *data,int a,int b){
	if( find(data,a)==find(data,b) ) return 1;
	return 0;
}