제출 #1073358

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1073358		pan	열쇠 (IOI21_keys)	C++17	100 / 100	500 ms	71232 KiB

keys

#include <bits/stdc++.h>
//#include "bits_stdc++.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) 
using namespace std;
mt19937 rng(chrono::system_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 unsigned long long ull;
typedef long double ld;
typedef pair<ld, ll> pd;
typedef pair<string, ll> psl;
typedef pair<ll, ll> pi;
typedef pair<pi, ll> pii;
typedef pair<pi, pi> piii;

ll par[300005];
ll find(ll x) {return ((par[x]==x)?x:par[x] = find(par[x]));}
void unite(ll from, ll to) {par[find(from)] = find(to);}

ll siz = LLONG_MAX;
vector<ll> minn, now,lock1;
vector<bool> visited;
vector<ll> onlock[300005];
ll havekey[300005], prevv[300005], key[300005], fin[300005];
vector<pi> adj[300005];

void del()
{
	for (ll u: now) havekey[key[u]] = 0;
	for (ll u: lock1) onlock[u].clear();
	now.clear(); lock1.clear();
}

void bfs(ll from, ll t)
{
	prevv[from] = t;
	queue<ll> q;
	q.push(from);
	while (q.size())
	{
		ll c = q.front();
		q.pop();
		if (find(c) != find(from))
		{
			// c will lead to better result
			// terminate
			unite(from,c);
			prevv[find(c)] = t;
			del();
			return;
			
		}
		if (visited[c]) continue;
		visited[c] = 1;
		now.pb(c);
		if (!havekey[key[c]])
		{
			havekey[key[c]] = 1;
			for (ll u: onlock[key[c]]) q.push(u);
			onlock[key[c]].clear();
		}
		for (pi u: adj[c])
		{
			if (havekey[u.s])
			{
				q.push(u.f);
			}
			else
			{
				lock1.pb(u.s);
				onlock[u.s].pb(u.f);
			}
		}
	}
	fin[from] = 1;
	if (siz > now.size())
	{
		siz = now.size();
		minn = now;
	}
	else if (siz == now.size())
	{
		for (ll u: now) minn.pb(u);
	}
	del();
}


std::vector<int> find_reachable(std::vector<int> r, std::vector<int> u, std::vector<int> v, std::vector<int> c)
{
	ll n = r.size(), m = u.size();
	for (ll i=0; i<m; ++i) havekey[i] = 0;
	for (ll i=0; i<n; ++i) par[i] = i, key[i]  = r[i], fin[i] = prevv[i] = 0;
	for (ll i=0; i<m; ++i)
	{
		adj[u[i]].pb(mp(v[i], c[i]));
		adj[v[i]].pb(mp(u[i], c[i]));
	}
	
	// bfs
	ll t = 1;
	bool done = 0;
	while (!done)
	{
		done = 1;
		visited.assign(n, 0);
		for (ll i=0; i<n; ++i)
		{
			if (find(i)==i && !fin[i] && prevv[i]!=t)
			{
				bfs(i, t);
				done = 0;
			}
		}
		t++;
	}
	vector<int> ans(n, 0);
	for (ll u: minn) ans[u] = 1;
	return ans;
}

컴파일 시 표준 에러 (stderr) 메시지

keys.cpp: In function 'void bfs(ll, ll)':
keys.cpp:98:10: warning: comparison of integer expressions of different signedness: 'll' {aka 'long long int'} and 'std::vector<long long int>::size_type' {aka 'long unsigned int'} [-Wsign-compare]
   98 |  if (siz > now.size())
      |      ~~~~^~~~~~~~~~~~
keys.cpp:103:15: warning: comparison of integer expressions of different signedness: 'll' {aka 'long long int'} and 'std::vector<long long int>::size_type' {aka 'long unsigned int'} [-Wsign-compare]
  103 |  else if (siz == now.size())
      |           ~~~~^~~~~~~~~~~~~

• Subtask 1: 9 / 9
• Subtask 2: 11 / 11
• Subtask 3: 17 / 17
• Subtask 4: 30 / 30
• Subtask 5: 33 / 33