Submission #1166351

#	Time	Username	Problem	Language	Result	Execution time	Memory
1166351		pan	Alternating Heights (CCO22_day1problem1)	C++20	25 / 25	133 ms	4680 KiB

Main

#include <bits/stdc++.h>
//#include "includeall.h"
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define endl '\n'
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define input(x) scanf("%lld", &x);
#define input2(x, y) scanf("%lld%lld", &x, &y);
#define input3(x, y, z) scanf("%lld%lld%lld", &x, &y, &z);
#define input4(x, y, z, a) scanf("%lld%lld%lld%lld", &x, &y, &z, &a);
#define print(x, y) printf("%lld%c", x, y);
#define show(x) cerr << #x << " is " << x << endl;
#define show2(x,y) cerr << #x << " is " << x << " " << #y << " is " << y << endl;
#define show3(x,y,z) cerr << #x << " is " << x << " " << #y << " is " << y << " " << #z << " is " << z << endl;
#define all(x) x.begin(), x.end()
#define discretize(x) sort(x.begin(), x.end()); x.erase(unique(x.begin(), x.end()), x.end());
#define FOR(i, x, n) for (ll i =x; i<=n; ++i) 
#define RFOR(i, x, n) for (ll i =x; i>=n; --i) 
#pragma GCC optimize("O3","unroll-loops")
using namespace std;
mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
//using namespace __gnu_pbds;
//#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
//#define ordered_multiset tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update>
typedef long long ll;
typedef long double ld;
typedef pair<ld, ll> pd;
typedef pair<string, ll> psl;
typedef pair<ll, ll> pi;
typedef pair<ll, pi> pii;
typedef pair<pi, pi> piii;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());


ll n, k, q;
vector<ll> adj[3005];
ll a[3005], vis[3005], ans[3005];
ll dfs(ll x)
{
	if (vis[x]==1) return 1;
	if (vis[x] == 2) return 0;
	vis[x] = 1;
	for (ll u: adj[x]) if (dfs(u)) return 1;
	vis[x] = 2;
	return 0;
}

void reset()
{
	for (ll i=1; i<=k; ++i) {adj[i].clear();}
}


bool cycle()
{
	for (ll i=1; i<=k; ++i) vis[i] = 0;
	for (ll i=1; i<=k; ++i) if (dfs(i)) return 1;
	return 0;
}

int main()
{
	cin.tie(nullptr);
	cout.tie(nullptr);
	ios::sync_with_stdio(false);
	cin >> n >> k >> q;
	for (ll i=0; i<n; ++i) cin >> a[i];
	ll idx = 0;
	for (ll i=0; i<n; i+=2)
	{
		reset();
		if (idx < i) idx = i;
		ll par = 0;
		for (ll j=i + 1; j<=idx; ++j)
		{
			if (par) adj[a[j-1]].pb(a[j]);
			else adj[a[j]].pb(a[j-1]);
			par^=1;
		}
		while (idx < n && !cycle()) 
		{
			idx++;
			//show(idx);
			if (par) adj[a[idx-1]].pb(a[idx]);
			else adj[a[idx]].pb(a[idx-1]);
			par^=1;
		}
		ans[i] = idx;
		//show2(i, idx);
	}
	idx = 1;
	for (ll i=1; i<n; i+=2)
	{
		reset();
		if (idx < i) idx = i;
		ll par = 0;
		for (ll j=i + 1; j<=idx; ++j)
		{
			if (par) adj[a[j-1]].pb(a[j]);
			else adj[a[j]].pb(a[j-1]);
			par^=1;
		}
		while (idx < n && !cycle()) 
		{
			idx++;
			//show(idx);
			if (par) adj[a[idx-1]].pb(a[idx]);
			else adj[a[idx]].pb(a[idx-1]);
			par^=1;
		}
		ans[i] = idx;
		//show2(i, idx);
	}
	while (q--)
	{
		ll x, y;
		cin >> x >> y; x--; y--;
		if (ans[x] > y) cout << "YES\n";
		else cout << "NO\n";
	}
	

	return 0;
}

• Subtask 1: 4 / 4
• Subtask 2: 6 / 6
• Subtask 3: 7 / 7
• Subtask 4: 8 / 8