제출 #1251216

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1251216		model_code	Migrations (IOI25_migrations)	C++20	65 / 100	295 ms	524 KiB

migrations

// solution/author-5.cpp
// {
//   "verdict": "partially_correct",
//   "except": {
//     "samples": "correct",
//     "from_root": "correct"
//   }
// }
// END HEADER
// ********************************************************
// ** Solution of Z = 5
// ********************************************************

#include <bits/stdc++.h>
#include "migrations.h"
using namespace std;

// variables for ahmed
namespace device {
	int N, B;
	vector<int> p;
	int u, v;
	int ubase, vbase;

	// initialization
	void init(int N_) {
		N = N_;
		B = int(ceil(log2(N)));
		p = { -1 };
		u = 0; v = 0;
		ubase = 0; vbase = 0;
	}

	// calculate distance
	int dist(int x, int y) {
		int d = 0;
		while (x != y) {
			if (x > y) {
				x = p[x];
			}
			else {
				y = p[y];
			}
			d += 1;
		}
		return d;
	}

	// function for ahmed
	int ahmed(int t, int parent) {
		// step #1. preparation
		p.push_back(parent);
		int d1 = dist(u, v);
		int d2 = dist(u, t);
		int d3 = dist(v, t);
		int pattern;
		if (d1 >= d2 && d1 >= d3) {
			pattern = 0; // Pattern A
		}
		else if (d1 <= d2 && d2 >= d3) {
			v = t;
			pattern = 1; // Pattern B
		}
		else {
			u = v;
			v = t;
			pattern = 2; // Pattern C
		}
		if (t == N - B * 2 - 1) {
			ubase = u;
			vbase = v;
		}

		// step #2. if pre-writing phase, don't write
		if (t < N - B * 2) {
			return 0;
		}

		// step #3. writing phase
		int index = t - (N - B * 2);
		int bit = (index < B ? (ubase >> index) & 1 : (vbase >> (index - B)) & 1);
		return pattern * 2 + bit;
	}
}

// dummy function for ahmed
int send_message(int N, int pos, int P) {
	if (pos == 1) {
		device::init(N);
	}
	int res = device::ahmed(pos, P);
	return res;
}

// function for hassan
std::pair<int, int> longest_path(vector<int> Cards) {
	// step #1. preparation
	int N = Cards.size();
	int B = int(ceil(log2(N)));

	// step #2. decoding of (u, v) in pre-writing phase
	int ubase = 0, vbase = 0;
	for (int i = 0; i < 2 * B; i++) {
		int t = i + (N - 2 * B);
		if (t >= 1) {
			int bit = (Cards[t] & 1);
			if (i < B) {
				ubase += bit << i;
			}
			else {
				vbase += bit << (i - B);
			}
		}
	}

	// step #3. decode patterns
	int u = ubase, v = vbase;
	for (int i = 0; i < 2 * B; i++) {
		int t = i + (N - 2 * B);
		if (t >= 1) {
			int pattern = Cards[t] / 2;
			if (pattern == 1) {
				v = t; // Pattern B
			}
			if (pattern == 2) {
				u = v;
				v = t; // Pattern C
			}
		}
	}

	return make_pair(u, v);
}

• Subtask 0: 0 / 0
• Subtask 1: 30 / 30
• Subtask 2: 35 / 70