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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <stdio.h>

using namespace std;
int uf[100000];
vector<pair<int,int> > ab;
vector<vector<int> > fr;
vector<vector<int> > br;
vector<int> usize;

int abc157d();
int unionFind(int);

int main()
{
    abc157d();
    return 0;
}

int abc157d(){
    int n,m,k;
    cin>>n>>m>>k;
    int i,j;
    fr=vector<vector<int> >(n);
    br=vector<vector<int> >(n);
    for(i=0;i<m;i++){
        int a,b;
        cin>>a>>b;
        a--;b--;
        ab.push_back(make_pair(a,b));
        fr[a].push_back(b);
        fr[b].push_back(a);
    }
    for(i=0;i<k;i++){
        int a,b;
        cin>>a>>b;
        a--;b--;
        br[a].push_back(b);
        br[b].push_back(a);
    }
    for(i=0;i<n;i++){
        uf[i]=i;
    }

    for(i=0;i<m;i++){
        int a= ab[i].first;
        int b= ab[i].second;
        if(uf[a]!=uf[b]){
            if(uf[a]>uf[b]){
                int tmp=a;
                a=b;
                b=tmp;
            }
            uf[uf[b]]=uf[a];
            unionFind(b);
        }
    }
    for(i=0;i<n;i++){
        unionFind(i);
    }

    usize = vector<int>(n,0);
    for(i=0;i<n;i++){
        usize[uf[i]]++;
    }
    for(i=0;i<n;i++){
        //printf("%d:%d size=%d\n",i,uf[i], usize[i]);
    }

    for(i=0;i<n;i++){
        int ans=usize[uf[i]] - 1;
        for(j=0;j<fr[i].size();j++){
            if(uf[fr[i][j]]==uf[i])ans--;
        }
        for(j=0;j<br[i].size();j++){
            if(uf[br[i][j]]==uf[i])ans--;
        }
        cout<<(i==0?"":" ")<<ans;
    }
    cout<<endl;

    return 0;
}

int unionFind(int x){
    if(uf[x]==x) return x;
    return uf[x] = unionFind(uf[x]);
}