Submission #1139157

#	Time	Username	Problem	Language	Result	Execution time	Memory
1139157		monkey133	Colors (RMI18_colors)	C++20	0 / 100	155 ms	8788 KiB

colors

#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, m;
vector<ll> adj[150005];
pi a[100005], b[100005];
ll par[100005], aa[100005];
set<ll> cc[100005];
ll open[100005];

ll find(ll x) {return ((par[x]==x)?x:par[x] = find(par[x]));}
void unite(ll x, ll y)
{
	x = find(x), y = find(y);
	if (x==y) return;
	if (cc[x].size() < cc[y].size()) swap(x, y); 
	par[y] = x;
	for (ll u: cc[y]) cc[x].insert(u);
}

void solve()
{
	cin >> n >> m;
	for (ll i=1; i<=n; ++i) par[i] = i, a[i].s = i, b[i].s = i, open[i] = 0;
	for (ll i=1; i<=n; ++i) cin >> aa[i];
	for (ll i=1; i<=n; ++i) {cin >> b[i].f;  a[i].f = b[i].f;}
	for (ll i=1; i<=n; ++i) {cc[i].clear(); adj[i].clear();}
	for (ll i=1; i<=n; ++i) cc[i].insert(aa[i]);
	for (ll i=1; i<=n; ++i) if (b[i].f > aa[i]) {cout << "0\n"; return;}
	for (ll i=0; i<m; ++i)
	{
		ll u,v;
		cin >> u >> v;
		adj[u].pb(v);
		adj[v].pb(u);
	}
	sort(a+1, a+n+1); sort(b+1, b+n+1);
	ll idx  =1;
	for (ll i=1; i<=n; ++i)
	{
		while (idx  <=n && a[idx].f <= b[i].f)
		{
			open[a[idx].s] = 1;
			for (ll u: adj[a[idx].s]) if (open[u]) unite(u, a[idx].s);
			idx++;
		}
		ll  x = find(b[i].s);
		if (cc[x].find(b[i].f)==cc[x].end()) {cout << "0\n"; return;}
	}
	cout << "1\n" << endl;
}

int main()
{
	ll t;
	cin >> t;
	while (t--) solve();
	return 0;
}

• Subtask 1: 0 / 15
• Subtask 2: 0 / 7
• Subtask 3: 0 / 8
• Subtask 4: 0 / 15
• Subtask 5: 0 / 7
• Subtask 6: 0 / 16
• Subtask 7: 0 / 10
• Subtask 8: 0 / 22