제출 #363483

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
363483		Return_0	동기화 (JOI13_synchronization)	C++17	50 / 100	335 ms	28780 KiB

synchronization

// #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);
#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.empty())return out<<"[]";out<< '[' << a[0]; for (int i = 0; ++i < (int)(a.size());) out<< ", " << a[i];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 = 2e5 + 7;

ll dp [N], up [N], tin [N], rtin [N], tout [N], T, n, m, q;
vector <ll> adj [N];
bool mark [N];
pll e [N];

void dfs (cl &v = 1, cl &pr = 1) {
	tin[v] = ++T;
	rtin[T] = v;
	for (auto &u : adj[v]) if (u ^ pr)	dfs(u, v);
	tout[v] = T;
}

struct seg {
	ll mx, l, r;
	seg *lt, *rt;
	inline seg (cl &L = 1, cl &R = N - 2) : l(L), r(R), mx(0) {}

	void build () {
		if (l == r) {
			mx = tout[rtin[l]];
			return;
		}
		lt = new seg (l, mid);
		rt = new seg (mid + 1, r);
		lt->build();
		rt->build();
		mx = (lt->mx, rt->mx);
	}

	void upd (cl &pos, cl &x) {
		if (l == r)	{
			mx = x;
			return;
		}
		if (pos <= mid)	lt->upd(pos, x);
		else rt->upd(pos, x);
		mx = max(lt->mx, rt->mx);
	}

	ll ask (cl &qr, cl &x) {
		if (mx < x)	return 1;
		if (l == r)	return l;
		if (r <= qr) {
			if (rt->mx >= x)	return rt->ask(qr, x);
			return lt->ask(qr, x);
		}
		ll y = 1;
		if (rt->mx >= x && rt->l <= qr)	y = rt->ask(qr, x);
		if (y == 1)	y = lt->ask(qr, x);
		return y;
	}
} mytree;

ll ask (cl &qr, cl &x) {
	ll ret = mytree.ask(qr, x);
	ret = rtin[ret];
	return ret;
}

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

	fill(dp, dp + N, 1);

	cin>> n >> m >> q;
	for (ll i = 1, v, u; i < n; i++) {
		cin>> v >> u;
		e[i] = {v, u};
		adj[v].push_back(u);
		adj[u].push_back(v);
	}

	dfs();
	mytree.r = n;	mytree.build();

	for (ll i = 0, x; i < m; i++) {
		cin>> x;
		if (tin[e[x].fr] > tin[e[x].sc])	swap(e[x].fr, e[x].sc);
		if (!mark[x])	{
			dp[ask(tin[e[x].fr], tout[e[x].fr])] += dp[e[x].sc] - up[e[x].sc];
			mytree.upd(tin[e[x].sc], 0);
		} else {
			dp[e[x].sc] = up[e[x].sc] = dp[ask(tin[e[x].sc], tout[e[x].sc])];
			mytree.upd(tin[e[x].sc], tout[e[x].sc]);
		}
		mark[x] ^= 1;
	}

	for (ll v; q--;) {
		cin>> v;
		cout<< dp[ask(tin[v], tout[v])] << '\n';
	}

	return 0;
}
/*

*/

컴파일 시 표준 에러 (stderr) 메시지

synchronization.cpp: In function 'std::ostream& operator<<(std::ostream&, const std::vector<_Tp>&)':
synchronization.cpp:44:2: warning: this 'if' clause does not guard... [-Wmisleading-indentation]
   44 |  if(a.empty())return out<<"[]";out<< '[' << a[0]; for (int i = 0; ++i < (int)(a.size());) out<< ", " << a[i];out<< ']';  }
      |  ^~
synchronization.cpp:44:32: note: ...this statement, but the latter is misleadingly indented as if it were guarded by the 'if'
   44 |  if(a.empty())return out<<"[]";out<< '[' << a[0]; for (int i = 0; ++i < (int)(a.size());) out<< ", " << a[i];out<< ']';  }
      |                                ^~~
synchronization.cpp: In constructor 'seg::seg(const ll&, const ll&)':
synchronization.cpp:62:12: warning: 'seg::r' will be initialized after [-Wreorder]
   62 |  ll mx, l, r;
      |            ^
synchronization.cpp:62:5: warning:   'll seg::mx' [-Wreorder]
   62 |  ll mx, l, r;
      |     ^~
synchronization.cpp:64:9: warning:   when initialized here [-Wreorder]
   64 |  inline seg (cl &L = 1, cl &R = N - 2) : l(L), r(R), mx(0) {}
      |         ^~~
synchronization.cpp: In member function 'void seg::build()':
synchronization.cpp:75:13: warning: left operand of comma operator has no effect [-Wunused-value]
   75 |   mx = (lt->mx, rt->mx);
      |         ~~~~^~

• Subtask 1: 0 / 10
• Subtask 2: 20 / 20
• Subtask 3: 10 / 10
• Subtask 4: 20 / 20
• Subtask 5: 0 / 40