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: Accepted)


// Problem : D - Friend Suggestions
// Contest : AtCoder - AtCoder Beginner Contest 157
// URL : https://atcoder.jp/contests/abc157/tasks/abc157_d
// Memory Limit : 1024 MB
// Time Limit : 2000 ms
// Powered by CP Editor (https://github.com/cpeditor/cpeditor)

#include<bits/stdc++.h>
using namespace std; 
#define ll long long int
#define pb push_back
#define mp make_pair
#define pll pair<ll,ll>
#define INF LLONG_MAX
#define PI acos(-1)
const ll MOD = 1e9 + 7;
#define fir first
#define sec second
#define all(x) x.begin(),x.end()
#define mem(a,b) memset(a,b,sizeof(a))
#define rep(c,a,b) for(ll c=a;c<b;c++)
#define IOS ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define min_heap priority_queue<pll,vector<pll>,greater<pll> >
ll dx4[]={-1,1,0,0};
ll dy4[]={0,0,-1,1};
ll dx8[]={-1,-1,-1,0,1,1,1,0};
ll dy8[]={-1,0,1,1,1,0,-1,-1};
bool isvalid(ll x,ll y,ll n,ll m)
{
    return (x>=0 and y>=0 and x<n and y<m);
}
ll isprime[1000];
vector<ll> prime;
void sievetoprime(ll n)
{
    mem(isprime,true);
    for(ll p=2;p*p<=n;p++)
    {
        if(isprime[p]==true)
        {
            for(ll i=p*p;i<=n;i+=p)
                isprime[i]=false;
        }
    }
}
ll lcm(ll a,ll b)
{
    return (a/__gcd(a,b))*b;
}
ll ind=0;
vector<ll> vis,leader;
vector<vector<ll> > adj,adj2;
vector<ll> comp[100005];
ll n,m,k;
void dfs(ll u)
{
	vis[u]=1;
	comp[ind].pb(u);
	for(auto i:adj[u])
	{
		if(vis[i]==0)
			dfs(i);
	}
}
void solve()
{
    cin>>n>>m>>k;
    ll a,b;
    adj.resize(n+1);
    leader.resize(n+1);
    adj2.resize(n+1);
    for(ll i=0;i<m;i++)
    {
    	cin>>a>>b;
    	adj[a].pb(b);
    	adj[b].pb(a);
    }
    vector<pll> v;
    for(ll i=0;i<k;i++)
    {
    	cin>>a>>b;
    	adj2[a].pb(b);
    	adj2[b].pb(a);
    	v.pb({a,b});
    }
    vis.resize(n+1,0);
    for(ll i=1;i<=n;i++)
    {
    	leader[i]=i;
    	if(vis[i]==0)
    	{
    		dfs(i);
    		ind++;
    	}
    }
    vector<ll> cnt(n+1,0);
    for(ll i=0;i<ind;i++)
    {
    	for(ll j=0;j<comp[i].size();j++)
    	{
    		cnt[comp[i][j]]=comp[i].size();
    		leader[comp[i][j]]=comp[i][0];
    	}
    }
    for(auto i:v)
    {
    	a=i.fir;
    	b=i.sec;
    	if(leader[a]==leader[b])
    	{
    		cnt[a]--;
    		cnt[b]--;
    	}
    }
    for(ll i=1;i<=n;i++)	
    	cout<<cnt[i]-1-adj[i].size()<<" ";
}
int main()
{
    ll inpoup = 0;
    if(inpoup==1)
    {
        #ifndef ONLINE_JUDGE
            freopen("input.txt","r",stdin);
            freopen("output.txt","w",stdout);
        #endif
    }
    IOS
    ll t;
    t=1;
    //cin>>t;
    while(t--)
    {
        solve();
    }
}