Submission #1030803

#	Time	Username	Problem	Language	Result	Execution time	Memory
1030803		fv3	Split the Attractions (IOI19_split)	C++14	22 / 100	592 ms	1048576 KiB

split

#include "split.h"
#include <bits/stdc++.h>

using namespace std;

vector<vector<pair<int, int>>> adj;
vector<int> sz, label;

pair<int, int> g[3];
int N;

int DFS(int index, int last)
{
	sz[index] = 1;
	for (auto&node:adj[index])
	{
		if (node.second == last) continue;
		node.first = DFS(node.second, index);
		sz[index] += node.first;
	}

	for (auto&node : adj[index])
	{
		if (node.second != last) continue;
		node.first = N - sz[index];
		break;
	}

	return sz[index];
}

int lbl_a = 0;
void labelA(int index, int last)
{
	if (lbl_a < g[0].first)
		label[index] = g[0].second;
	else
		label[index] = g[2].second;

	lbl_a++;
	for (auto node : adj[index])
	{
		if (node.second == last) continue;
		labelA(node.second, index);
	}
}

int lbl_b = 0;
void labelB(int index, int last)
{
	if (lbl_b < g[1].first)
		label[index] = g[1].second;
	else
		label[index] = g[2].second;

	lbl_b++;
	for (auto node : adj[index])
	{
		if (node.second == last) continue;
		labelB(node.second, index);
	}
}


vector<int> find_split(int n, int a, int b, int c, vector<int> p, vector<int> q)
{
	N = n;
	g[0] = {a, 1}; g[1] = {b, 2}; g[2] = {c, 3};
	sort(g, g + 3);

	const int M = p.size();
	adj = vector<vector<pair<int, int>>>(N);
	sz = label = vector<int>(N);

	for (int i = 0; i < M; i++)
	{
		adj[p[i]].push_back({0, q[i]});
		adj[q[i]].push_back({0, p[i]});
	}

	// Let a be the smallest
	// if sz[i] >= a and (N - sz[i]) >= min(b, c) or opposite it can be done

	// Find tree sizes
	DFS(0, 0);


	for (int i = 0; i < N; i++)
	{
		sort(adj[i].begin(), adj[i].end());

		// Find the smallest subtree larger than g[0].first
		if (adj[i].back().first < g[0].first) continue;

		int l = 0, r = adj[i].size() - 1;
		while (l < r)
		{
			int c = (l + r) / 2;
			if (adj[i][c].first < g[0].first)
				l = c + 1;
			else
				r = c;
		}
		int subtree = adj[i][l].first; 
		if (sz[i] - subtree < g[1].first) continue;

		labelA(adj[i][l].second, i);
		labelB(i, adj[i][l].second);

		break;
	}
	
	return label;
}

• Subtask 1: 0 / 7
• Subtask 2: 0 / 11
• Subtask 3: 22 / 22
• Subtask 4: 0 / 24
• Subtask 5: 0 / 36