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  1.  
  2.     #include <bits/stdc++.h>
  3.  
  4.     using namespace std;
  5.  
  6.     using ll = long long;
  7.     const int MOD = 1000000007; 
  8.     const int MOD2 =  998244353; 
  9.     const ll INF = 1e18;
  10.     const int MX = 1000001; //check the limits, dummy
  11.  
  12.  
  13.     ll modExp(ll base, ll power) {
  14.         if (power == 0) {
  15.             return 1;
  16.         } else {
  17.             ll cur = modExp(base, power / 2); cur = cur * cur; cur = cur % MOD;
  18.             if (power % 2 == 1) cur = cur * base;
  19.             cur = cur % MOD;
  20.             return cur;
  21.         }
  22.     }
  23.  
  24.     ll inv(ll base) {
  25.         return modExp(base, MOD-2);
  26.     }
  27.  
  28.  
  29.     ll mul(ll A, ll B) {
  30.         return (A*B)%MOD;
  31.     }
  32.  
  33.     ll add(ll A, ll B) {
  34.         return (A+B)%MOD;
  35.     }
  36.  
  37.     ll dvd(ll A, ll B) {
  38.         return mul(A, inv(B));
  39.     }
  40.  
  41.     ll sub(ll A, ll B) {
  42.         return (A-B+MOD)%MOD;
  43.     }
  44.     ll cielDiv(ll A , ll B) {
  45.         return (A + B - 1)/B;
  46.     } 
  47.  
  48.     ll* facs = new ll[MX];
  49.     ll* facInvs = new ll[MX];
  50.  
  51.     ll choose(ll a, ll b) {
  52.         if (b > a) return 0;
  53.         if (a < 0) return 0;
  54.         if (b < 0) return 0;
  55.         ll cur = facs[a];
  56.         cur = mul(cur, facInvs[b]);
  57.         cur = mul(cur, facInvs[a-b]);
  58.         return cur;
  59.     }
  60.  
  61.     void initFacs() {
  62.         facs[0] = 1; 
  63.         facInvs[0] = 1;
  64.         for (int i = 1 ; i < MX ; i ++ ) {
  65.             facs[i] = (facs[i-1] * i) % MOD;
  66.             facInvs[i] = inv(facs[i]);
  67.         }
  68.     }
  69.     ll n, m,k ;
  70.     const int maxn = 200005;
  71.     vector<ll> adj[maxn];
  72.     bool seen[maxn]; 
  73.     vector<ll> eulers;
  74.     void dfs(ll a, ll p) {
  75.         seen[a] = true; 
  76.         eulers.push_back(a);        
  77.         for (ll c : adj[a]) {
  78.             if (!seen[c]) {
  79.                 dfs(c,a); 
  80.             }
  81.         }
  82.         eulers.push_back(a); 
  83.     }
  84.  
  85.  
  86.  
  87.  
  88.     int main()  {
  89.         ios_base::sync_with_stdio(0); cin.tie(0);  
  90.         cin >> n >> m >> k ;
  91.         for (int i = 0 ;i < m ; i ++) {
  92.             ll a , b; cin >> a >> b; 
  93.             adj[a].push_back(b);
  94.             adj[b].push_back(a);
  95.         }
  96.         dfs(1,0); 
  97.         // 1) correct ceiling
  98.         ll limit = ( (ll)eulers.size() + k - 1 ) / k;
  99.  
  100.         // 2) print exactly k lines
  101.         int printed = 0, p = 0;
  102.         while (printed < k) {
  103.             if (p < (int)eulers.size()) {
  104.                 int sz = min<ll>(limit, eulers.size() - p);
  105.                 cout << sz;
  106.                 for (int i = 0; i < sz; i++, p++) {
  107.                     cout << ' ' << eulers[p];
  108.                 }
  109.                 cout << "\n";
  110.             } else {
  111.                 // pad with a trivial single‐node tour
  112.                 cout << "1 1\n";
  113.             }
  114.             printed++;
  115.         }   
  116.  
  117.         return 0;
  118.     }
  119.  
Success #stdin #stdout 0.01s 8280KB
comments (?)
stdin
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5 4 2
1 2
1 3
1 4
1 5
stdout
copy 
5 1 2 2 3 3
5 4 4 5 5 1
https://ideone.com/3SNVDF
language:
C++ (gcc 8.3)
created:
10 days ago
visibility:
public

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