Submission #335766

#TimeUsernameProblemLanguageResultExecution timeMemory
335766534351Praktični (COCI18_prakticni)C++17
26 / 130
140 ms16108 KiB
#include <bits/stdc++.h>
// #include <ext/pb_ds/assoc_container.hpp>

using namespace std;
// using namespace __gnu_pbds;

// template<class T>
// using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

template<class T, class U>
void ckmin(T &a, U b)
{
    if (a > b) a = b;
}

template<class T, class U>
void ckmax(T &a, U b)
{
    if (a < b) a = b;
}

#define MP make_pair
#define PB push_back
#define LB lower_bound
#define UB upper_bound
#define fi first
#define se second
#define SZ(x) ((int) (x).size())
#define ALL(x) (x).begin(), (x).end()
#define FOR(i, a, b) for (auto i = (a); i < (b); i++)
#define FORD(i, a, b) for (auto i = (a) - 1; i >= (b); i--)

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpi;
typedef vector<pll> vpl;
typedef pair<pii, int> ppi;

const int MAXN = 100013;

int N, M, ans;
vector<ppi> edge[MAXN];
bitset<MAXN> seen;
int parent[MAXN], val[MAXN], depth[MAXN];
vi guys[32]; //set of edge ids that need to be xored by 2^i.
int res[32], group[32];

void dfs(int u)
{
    seen[u] = true;
    for (auto e : edge[u])
    {
        int v = e.se;
        if (seen[v]) continue;
        parent[v] = u;
        depth[v] = depth[u] + 1;
        val[v] = (val[u] ^ e.fi.fi);
        dfs(v);
    }
}

int32_t main()
{
    cout << fixed << setprecision(12);
    cerr << fixed << setprecision(4);
    ios_base::sync_with_stdio(false); cin.tie(0);
    cin >> N >> M;
    FOR(i, 0, M)
    {
        int u, v, p;
        cin >> u >> v >> p;
        u--; v--;
        edge[u].PB({{p, i}, v});
        edge[v].PB({{p, i}, u});
    }
    parent[0] = N;
    dfs(0);
    FOR(u, 0, N)
    {
        for (auto e : edge[u])
        {
            int v = e.se;
            if (v == parent[u] || depth[v] >= depth[u]) continue;
            int c = (val[u] ^ val[v] ^ e.fi.fi);
            FORD(j, 31, 0)
            {
                if (c & (1 << j))
                {
                    guys[j].PB(e.fi.se);
                    // cerr << j << ' ' << e.fi.se << endl;
                }
            }
        }
    }
    int done = 0;
    FOR(i, 0, 31)
    {
        if (guys[i].empty()) continue;
        if (done & (1 << i)) continue;
        done |= (1 << i);
        res[ans] = (1 << i);
        group[ans] = i;
        FOR(j, i + 1, 31)
        {
            if (guys[i] == guys[j])
            {
                res[ans] |= (1 << j);
                done |= (1 << j);
            }
        }
        ans++;
    }
    cout << ans << '\n';
    FOR(i, 0, ans)
    {
        cout << res[i] << ' ' << SZ(guys[group[i]]);
        for (int x : guys[group[i]])
        {
            cout << ' ' << x + 1;
        }
        cout << '\n';
    }
    return 0;
}
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