제출 #349514

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
349514		Return_0	Bitaro’s Party (JOI18_bitaro)	C++17	100 / 100	1312 ms	134972 KiB

bitaro

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
 
// #pragma GCC optimize ("Ofast,unroll-loops")
// #pragma GCC target ("avx,avx2,fma")
 
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);
#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) {
	if(!a.size())return cout<<"[]";cout<< '[' << a[0]; for (int i = 0; ++i < (int)(a.size());) cout<< ", " << a[i];cout<< ']';  }
template <class T, typename _t = less <T> > using Tree = tree <T, null_type, _t, rb_tree_tag, tree_order_statistics_node_update>;
 
cl N = 1e5 + 7, SQ = 150;

ll dis [N], n, m, q;
vector <ll> adj [N], jda [N], vec;
pll dp [N][SQ + 2], tmp [SQ + 2];
bool mark [N];

ll Naive (ll st, ll sz) {
	ll ans = -1;
	memset(dis, - 1, sizeof dis);
	vec.clear();
	for (ll i = 0, x; i < sz; i++) {
		cin>> x;
		vec.push_back(x);
	}
	for (auto &v : vec)	mark[v] = true;
	dis[st] = 0;
	for (ll v = st; v >= 1; v--) {
		for (auto u : adj[v])	if (~dis[u])	dis[v] = max(dis[v], dis[u] + 1);
		if (!mark[v])	ans = max(ans, dis[v]);
	}
	for (auto &v : vec)	mark[v] = false;
	return ans;
}

ll solve (ll v, ll sz) {
	vec.clear();
	for (ll i = 0, x; i < sz; i++) {
		cin>> x;
		vec.push_back(x);
		mark[x] = 1;
	}

	ll ans = -1;
	for (ll i = 0; i <= SQ; i++)	if (dp[v][i].sc > 0 && !mark[dp[v][i].sc])	ans = max(ans, dp[v][i].fr);
	
	for (auto &x : vec)	mark[x] = 0;

	return ans;
}

void _merge (ll v, ll u) {
	ll p1 = 0, p2 = 0, tot = 0;

	while (tot <= SQ && dp[v][p1].sc && dp[u][p2].sc) {
		if (mark[dp[v][p1].sc]) {
			p1++; continue;
		}
		if (mark[dp[u][p2].sc]) {
			p2++; continue;
		}

		if (dp[v][p1].fr >= dp[u][p2].fr + 1) {
			mark[dp[v][p1].sc] = 1;
			tmp[tot++] = dp[v][p1++];
		}

		else {
			mark[dp[u][p2].sc] = 1;
			tmp[tot++] = {dp[u][p2].fr + 1, dp[u][p2].sc};
			p2++;
		}
	}

	while (tot <= SQ && dp[v][p1].sc) {
		if (mark[dp[v][p1].sc]) {
			p1++; continue;
		}
		mark[dp[v][p1].sc] = 1;
		tmp[tot++] = dp[v][p1++];
	}

	while (tot <= SQ && dp[u][p2].sc) {
		if (mark[dp[u][p2].sc]) {
			p2++; continue;
		}
		mark[dp[u][p2].sc] = 1;
		tmp[tot++] = {dp[u][p2].fr + 1, dp[u][p2].sc};
		p2++;
	}

	for (ll i = 0; i <= tot; i++)	mark[tmp[i].sc] = 0, dp[v][i] = tmp[i], tmp[i] = {0, 0};
}

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

	cin>> n >> m >> q;
	for (ll i = 0, v, u; i < m; i++) cin>> v >> u, adj[v].push_back(u), jda[u].push_back(v);

	for (ll v = 1; v <= n; v++) {
		dp[v][0] = {0, v};
		for (auto &u : jda[v])	_merge(v, u);
	}

	for (ll i = 0, t, v; i < q; i++) {
		cin>> v >> t;
		if (t < SQ) cout<< solve(v, t) << '\n';
		else cout<< Naive(v, t) << '\n';
	}

	return 0;
}
/*

*/

컴파일 시 표준 에러 (stderr) 메시지

bitaro.cpp: In function 'std::ostream& operator<<(std::ostream&, const std::vector<_Tp>&)':
bitaro.cpp:43:2: warning: this 'if' clause does not guard... [-Wmisleading-indentation]
   43 |  if(!a.size())return cout<<"[]";cout<< '[' << a[0]; for (int i = 0; ++i < (int)(a.size());) cout<< ", " << a[i];cout<< ']';  }
      |  ^~
bitaro.cpp:43:33: note: ...this statement, but the latter is misleadingly indented as if it were guarded by the 'if'
   43 |  if(!a.size())return cout<<"[]";cout<< '[' << a[0]; for (int i = 0; ++i < (int)(a.size());) cout<< ", " << a[i];cout<< ']';  }
      |                                 ^~~~

• Subtask 1: 7 / 7
• Subtask 2: 7 / 7
• Subtask 3: 86 / 86