Submission #382246

#	Time	Username	Problem	Language	Result	Execution time	Memory
382246		Return_0	Rigged Roads (NOI19_riggedroads)	C++17	100 / 100	1295 ms	93028 KiB

riggedroads

#pragma GCC optimize("Ofast,unroll-loops")
// #pragma comment(linker, "/stack:200000000")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma,tune=native")

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

typedef int ll;
typedef long double ld;
typedef pair <ll, ll> pll;

#ifdef SINA
#define dbg(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1> void __f(const char* name, Arg1&& arg1) { cout << name << " : " << arg1 << std::endl; }
template <typename Arg1, typename... Args> void __f (const char* names, Arg1&& arg1, Args&&... args) {
	const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...); }
#define dbg2(x, j, n) cout<< #x << " : "; output((j), (n), x, 1); cout.flush();
#else
#define dbg(...) 0
#define dbg2(x, j, n) 0
#endif
#define SZ(x) ((ll)((x).size()))
#define File(s, t) freopen(s ".txt", "r", stdin); freopen(t ".txt", "w", stdout);
#define input(j, n, a) for (int _i = (j); _i < (n)+(j); _i++) cin>> a[_i];
#define output(j, n, a, t) for (int _i = (j); _i < (n)+(j); _i++) cout<< a[_i] << (((t) && _i != (n)+(j)-1)? ' ' : '\n');
#define kill(x) return cout<< (x) << endl, 0
#define cl const ll
#define fr first
#define sc second
#define lc (v << 1)
#define rc (lc | 1)
#define mid ((l + r) >> 1)
#define All(x) (x).begin(), (x).end()

cl inf = sizeof(ll) == 4 ? (1e9 + 10) : (3e18), mod = 1e9 + 7, MOD = 998244353;

template <class A,class B> ostream& operator << (ostream& out,const pair<A,B>&a){return out<<'('<<a.first<<", "<<a.second<<')';}
template <class A> ostream& operator << (ostream& out, const vector<A> &a) {
	out<< '['; for (int i = -1; ++i < int(a.size());) out<< a[i] << (i + 1 < int(a.size()) ? ", " : ""); return out<<']'; }
template <class T, typename _t = less <T> > using Tree = tree <T, null_type, _t, rb_tree_tag, tree_order_statistics_node_update>;

cl N = 3e5 + 7, lg = 19;

// struct myHash { ll operator () (pll x) const { return 101 * x.fr + x.sc; } };

ll P [N], par [N], anc [N][lg], h [N], lst, n, m;
pll E [N];
// gp_hash_table <pll, ll, myHash> mp;
map <pll, ll> mp;
vector <ll> adj [N], vec;

void dfs (cl &v = 1, cl &pr = 1) {
	h[v] = h[pr] + 1;
	anc[v][0] = pr;
	for (ll i = 1; i < lg; i++) anc[v][i] = anc[anc[v][i - 1]][i - 1];
	for (auto &u : adj[v]) if (u != pr) dfs(u, v);
}

inline ll up (ll v, cl &d) { for (ll i = lg; i--;) if (d >> i & 1) v = anc[v][i]; return v; }
inline ll LCA (ll v, ll u) {
	if (h[v] > h[u]) swap(v, u);
	u = up(u, h[u] - h[v]);
	if (v == u) return v;
	for (ll i = lg; i--;) if (anc[v][i] ^ anc[u][i]) v = anc[v][i], u = anc[u][i];
	return anc[v][0];
}

ll getpar (cl &v) { return par[v] = (par[v] == v ? v : getpar(par[v])); }

void ok (ll v, ll u) {
	ll lca = LCA(v, u);
	vec.clear();

	for (v = getpar(v); h[lca] < h[v]; v = par[v]) {
		vec.push_back(mp[{v, anc[v][0]}]);
		par[v] = getpar(anc[v][0]);
	}

	for (u = getpar(u); h[lca] < h[u]; u = par[u]) {
		vec.push_back(mp[{u, anc[u][0]}]);
		par[u] = getpar(anc[u][0]);
	}

	sort(All(vec));
	for (auto &x : vec) P[x] = ++lst;
}
inline void fuck (pll &x) { if (h[x.fr] > h[x.sc]) swap(x.fr, x.sc); par[x.sc] = getpar(x.fr); }

int main ()
{
	ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);

	iota(par, par + N, 0);
	cin>> n >> m;
	for (ll i = 1; i <= m; i++) cin>> E[i].fr >> E[i].sc;
	for (ll i = 1, x; i < n; i++) {
		cin>> x;
		mp[E[x]] = mp[{E[x].sc, E[x].fr}] = x;
		adj[E[x].fr].push_back(E[x].sc);
		adj[E[x].sc].push_back(E[x].fr);
	}

	dfs();

	for (ll i = 1; i <= m; i++) if (!P[i]) {
		if (mp.find(E[i]) == mp.end()) ok(E[i].fr, E[i].sc);
		else fuck(E[i]);
		P[i] = ++lst;
	}

	output(1, m, P, 1);

	cerr<< "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";

	return 0;
}
/*
4 5
3 4
1 2
2 3
1 3
1 4
2 4 5

4 4
1 2
1 4
2 3
3 4
1 3 4

*/

• Subtask 1: 8 / 8
• Subtask 2: 19 / 19
• Subtask 3: 9 / 9
• Subtask 4: 10 / 10
• Subtask 5: 10 / 10
• Subtask 6: 12 / 12
• Subtask 7: 32 / 32