Submission #603566

#TimeUsernameProblemLanguageResultExecution timeMemory
603566SeDunionSplit the Attractions (IOI19_split)C++17
0 / 100
4 ms4948 KiB
#include "split.h"
#include<algorithm>
#include<iostream>
#include<vector>
#include<assert.h>

using namespace std;
const int N = 2e5 + 123;

vector<int>g[N];

int used[N];

int n, a, b, c;

vector<int>vec, ans;

int av = -1, bv = -1;

int tin[N], tout[N], timer, sz[N];

void dfs(int v) {
	used[v] = 1;
	sz[v] = 1;
	tin[v] = timer++;
	for (int to : g[v]) if (!used[to]) {
		dfs(to);
		sz[v] += sz[to];
		if (sz[to] >= a && n - sz[to] >= b) {
			av = to, bv = v;
		}
		if (sz[to] >= b && n - sz[to] >= a) {
			av = v, bv = -to;
		}
	}
	tout[v] = timer++;
}

bool cmp(int a, int b) {
	return tin[a] < tin[b];
}

bool upper(int a, int b) { return tin[a] <= tin[b] && tout[a] >= tout[b]; } 

int gg[4];

vector<int> find_split(int n_, int a_, int b_, int c_, vector<int> p, vector<int> q) {
	n = n_, a = a_, b = b_, c = c_;
	if (a > b) swap(a, b);
	if (b > c) swap(b, c);
	if (a > b) swap(a, b);
	if (a_ == a && b_ == b && c_ == c) gg[1] = 1, gg[2] = 2, gg[3] = 3;
	if (a_ == a && b_ == c && c_ == b) gg[1] = 1, gg[3] = 2, gg[2] = 3;
	if (a_ == b && b_ == a && c_ == c) gg[2] = 1, gg[1] = 2, gg[3] = 3;
	if (a_ == b && b_ == c && c_ == a) gg[2] = 1, gg[3] = 2, gg[1] = 3;
	if (a_ == c && b_ == a && c_ == b) gg[3] = 1, gg[1] = 2, gg[2] = 3;
	if (a_ == c && b_ == b && c_ == a) gg[3] = 1, gg[2] = 2, gg[1] = 3;
	// a <= b <= c
	// a <= n/3 ?
	// b <= n/2 ?
	int m = p.size();
	for (int i = 0 ; i < m ; ++ i) {
		g[p[i]].emplace_back(q[i]);
		g[q[i]].emplace_back(p[i]);
	}
	int root = 0;
	for (int i = 0 ; i < n ; ++ i) {
		if ((int)g[i].size() == 1) root = i;
	}
	ans = vector<int>(n, 3);
	dfs(root);
	if (av == -1) return vector<int>(n, 0);
	if (bv < 0) {
		bv = -bv;
		root = bv;
		dfs(root);
	}
	vector<int>vec;
	for (int i = 0 ; i < n ; ++ i) {
		vec.emplace_back(i);
	}
	sort(vec.begin(), vec.end(), cmp);
	for (int v : vec) {
		if (a>0 && upper(av, v)) {
			ans[v] = 1, a--;
		}
		if (b>0 && !upper(av, v)) {
			ans[v] = 2, b--;
		}
	}
	for (int &i : ans) i = gg[i];
	return ans;
}
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