1005F - Berland and the Shortest Paths - CodeForces Solution

#include <bits/stdc++.h>
using namespace std;

using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;

#define llmx LONG_LONG_MAX
#define llmn LONG_LONG_MIN
#define pb push_back
#define dbg(v)                                                                 \
  cout << "Line(" << __LINE__ << ") -> " << #v << " = " << (v) << '\n';

void dfs(vector<vector<int>> &incl, int i, int &k, vector<int> &cur) {
  if (k == 0)
    return;
  if (i == (int)incl.size()) {
    for (int a : cur) {
      cout << a;
    }
    cout << '\n';
    --k;
    return;
  }

  for (int opt : incl[i]) {
    cur[opt] = 1;
    dfs(incl, i + 1, k, cur);
    if (k == 0)
      return;
    cur[opt] = 0;
  }
}

void solve() {
  int n, m, k;
  cin >> n >> m >> k;

  vector<vector<pair<int, int>>> g(n);
  for (int i = 0; i < m; ++i) {
    int a, b;
    cin >> a >> b;
    g[a - 1].pb({b - 1, i});
    g[b - 1].pb({a - 1, i});
  }

  vector<vector<int>> incl(n);
  vector<int> mn_dist(n, INT_MAX);
  mn_dist[0] = 0;
  queue<int> q;
  q.push(0);
  int dist = 0;
  while (q.size()) {
    int size = q.size();
    for (int i = 0; i < size; ++i) {
      int cur = q.front();
      q.pop();

      for (auto &pr : g[cur]) {
        int nb = pr.first, edge_idx = pr.second;
        if (mn_dist[nb] < dist)
          continue;
        incl[nb].pb(edge_idx);
        if (mn_dist[nb] > dist) {
          q.push(nb);
          mn_dist[nb] = dist;
        }
      }
    }
    ++dist;
  }

  int total = 1;
  for (int i = 1; i < n && total < k; ++i) {
    total *= incl[i].size();
  }
  k = min(k, total);
  cout << k << '\n';

  vector<int> cur(m);
  dfs(incl, 1, k, cur);
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);

  solve();
  cout << '\n';

  return 0;
}