제출 #438643

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
438643		fleimgruber	던전 (IOI21_dungeons)	C++17	63 / 100	7115 ms	737016 KiB

dungeons

// this solves subtask 1, 3, 4, 5. the complexity is alright,
// but for larger n we get MLE. instead of binary lifting, we
// should probably use some other base, HLD, those linear jump pointers..
#include "dungeons.h"
#include <bits/stdc++.h>

using namespace std;

const int MAX_N = 50001; // subtasks 1, 3, 4, 5
const int MAX_LOG = 25; // 2^24 > 10^7

// dp[i][j][k] = we're at i, simulate for 2^j steps and
// win against an opponent if strength < 2^k (we don't gain strength)
struct {
	int end; // where do we end up at?
	long long gained; // total strength gained on the path
	// max(sum(gained) - opponent) along the simulation,
	// for all opponents we lose against
	long long max;
} dp[MAX_N][MAX_LOG][MAX_LOG];
int n, s[MAX_N], w[MAX_N];

// because p[i] = s[i], every time we lose, our strength doubles
struct Subtask2 {
	struct {
		int end;
		long long gained;
		long long min; // min(sum(gained) - opponent)
	} dp[400001][MAX_LOG];
	int l[400001];

	Subtask2(vector<int>& l_) {
		copy_n(l_.begin(), n, l);
		l[n] = n;
		for (int j = 0; j < MAX_LOG; j++)
			for (int i = 0; i <= n; i++) {
				auto& x = dp[i][j];
				if (j == 0)
					if (i == n) {
						x.end = n;
						x.gained = x.min = 0;
					} else {
						x.end = w[i];
						x.gained = s[i];
						x.min = -s[i];
					}
				else {
					auto& a = dp[i][j - 1];
					auto& b = dp[a.end][j - 1];
					x.end = b.end;
					x.gained = a.gained + b.gained;
					x.min = min(a.min, a.gained + b.min);

				}
			}
	}

	long long simulate(int at, int s_init) {
		long long strength = 0;
		auto lose = [&](int i, int j) {
			return strength + dp[i][j].min < 0;
		};
		while (at != n) {
			// jump until we lose
			for (int j = MAX_LOG - 1; j >= 0; j--)
				if (!lose(at, j)) {
					auto& x = dp[at][j];
					at = x.end;
					strength += x.gained;
				}
			// lose (if we're at the end, this adds 0)
			strength += s[at];
			at = l[at];
		}
		return strength;
	}
};
optional<Subtask2> sub2; // don't waste memory :)

void init(int n_, vector<int> s_, vector<int> p,
		vector<int> w_, vector<int> l) {
	n = n_;
	copy_n(s_.begin(), n, s);
	copy_n(w_.begin(), n, w);
	w[n] = n;
	if (n > 50000) {
		sub2.emplace(l);
		return;
	}
	for (int j = 0; j < MAX_LOG; j++)
		for (int i = 0; i <= n; i++)
			for (int k = 0; k < MAX_LOG; k++) {
				auto& x = dp[i][j][k];
				if (j == 0)
					if (i == n) {
						x.end = n;
						x.gained = x.max = 0;
					} else if (s[i] < (1 << k)) { // we win
						x.end = w[i];
						x.gained = s[i];
						x.max = LLONG_MIN;
					} else {
						x.end = l[i];
						x.gained = p[i];
						// 1 step behind (we didn't gain 'gained' yet)
						x.max = -s[i];
					}
				else {
					auto& a = dp[i][j - 1][k];
					auto& b = dp[a.end][j - 1][k];
					x.end = b.end;
					x.gained = a.gained + b.gained;
					x.max = max(a.max, a.gained + b.max);
				}
			}
}

long long simulate(int at, int s_init) {
	if (n > 50000)
		return sub2->simulate(at, s_init);
	long long strength = s_init;
	int k = 0;
	auto advance_k = [&] { // msb
		if (strength >= (1 << 24))
			k = 24;
		else
			while ((1 << (k + 1)) <= strength)
				k++;
	};
	// find first time our dp takes the wrong turn. once this happens,
	// the msb will change => happens only log(something) times
	// at this vertex we win:
	//	strength + gained - opponent >= 0
	// => strength + dp.max >= 0
	// but dp loses (dp.max is only calculated for losing edges)
	auto wrong_turn = [&](int i, int j, int k) {
		return strength + dp[i][j][k].max >= 0;
	};
	while (at != n) {
		advance_k();
		// jump to point where msb changes
		for (int j = MAX_LOG - 1; j >= 0; j--)
			if (!wrong_turn(at, j, k)) { // doesn't change => jump forward
				auto& x = dp[at][j][k];
				at = x.end;
				strength += x.gained;
			} // else it changes, so don't jump forward
		// now wrong_turn(at, 0, k) holds, so we win
		strength += s[at];
		at = w[at];
	}
	return strength;
}

• Subtask 1: 11 / 11
• Subtask 2: 0 / 26
• Subtask 3: 13 / 13
• Subtask 4: 12 / 12
• Subtask 5: 27 / 27
• Subtask 6: 0 / 11