https://www.acmicpc.net/problem/24444
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
#include <ext/rope>
#define fastio ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)
using namespace std;
using namespace __gnu_cxx;
#define X first
#define Y second
#define int int64_t
#define sz(v) (int)(v).size()
#define all(v) (v).begin(), (v).end()
#define rall(v) (v).rbegin(), (v).rend()
#define Compress(v) sort(all(v)), (v).erase(unique(all(v)), (v).end())
#define OOB(x, y) ((x) < 0 || (x) >= n || (y) < 0 || (y) >= m)
#define IDX(v, x) (lower_bound(all(v), x) - (v).begin())
#define debug(x) cout << (#x) << ": " << (x) << '\n'
using ll = long long;
using ull = unsigned long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using tii = tuple<int, int, int>;
template <typename T>
using wector = vector<vector<T>>;
vector<int> adj[100001];
int n, m, r;
vector<int> bfs(int st)
{
vector<int> dist(n + 1, -1), v(n + 1, 0);
queue<int> Q;
dist[st] = 0;
int idx = 1;
Q.push(st);
while (!Q.empty())
{
auto cur = Q.front();
Q.pop();
v[cur] = idx++;
for (auto &nxt : adj[cur])
{
if (dist[nxt] != -1 || nxt < 1 || nxt > n)
continue;
Q.push(nxt);
dist[nxt] = dist[cur] + 1;
}
}
return v;
}
int32_t main()
{
fastio;
cin >> n >> m >> r;
for (int i = 0; i < m; i++)
{
int u, v;
cin >> u >> v;
adj[u].push_back(v);
adj[v].push_back(u);
}
for (int i = 1; i <= n; i++)
sort(all(adj[i]));
auto dist = bfs(r);
for (int i = 1; i <= n; i++)
cout << (dist[i] == -1 ? 0 : dist[i]) << "\n";
}
BFS를 이용해 그래프의 정점들을 순회하면서 방문 순서를 기록해주면 됩니다.