제출 #1052005

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1052005		Tob	Split the Attractions (IOI19_split)	C++14	18 / 100	46 ms	13788 KiB

split

#include <bits/stdc++.h>

#include "split.h"

#define F first
#define S second
#define all(x) x.begin(), x.end()

using namespace std;

typedef long long ll;

const int N = 1e5 + 7;

int n, A, B, fo = -1, fox;
int siz[N];
bool bio[N];
vector <int> no;
vector <int> adj[N];

void ldfs(int x) {
	bio[x] = 1;
	no.push_back(x);
	for (auto y : adj[x]) {
		if (bio[y]) continue;
		ldfs(y);
	}
}

void dfs(int x) {
	bio[x] = 1;
	siz[x] = 1;
	vector <int> v;
	for (auto y : adj[x]) {
		if (bio[y]) continue;
		siz[x] += siz[y];
		v.push_back(siz[y]);
	}
	if (siz[x] != n) v.push_back(n-siz[x]);
	sort(all(v));
	if (v.back() >= B) {
		if (n-v.back() >= A) {
			fo = 0;
			fox = x;
		}
	}
	else {
		if (v.back() >= A) {
			fo = 1;
			fox = x;
		}
	}
}

vector<int> find_split(int n, int a, int b, int c, vector<int> p, vector<int> q) {
	n = a+b+c;
	int m = p.size();
	vector <int> res(n, 0);
	for (int i = 0; i < m; i++) {
		adj[p[i]].push_back(q[i]);
		adj[q[i]].push_back(p[i]);
	}
	int mxdeg = 0;
	for (int i = 0; i < n; i++) mxdeg = max(mxdeg, (int)adj[i].size());
	if (mxdeg <= 2) {
		ldfs(0);
		for (int i = 0; i < a; i++) res[no[i]] = 1;
		for (int i = a; i < a+b; i++) res[no[i]] = 2;
		for (int i = a+b; i < n; i++) res[no[i]] = 3;
	}
	else if (a == 1) {
		ldfs(0);
		for (int i = 0; i < b; i++) res[no[i]] = 2;
		for (int i = b; i < n-1; i++) res[no[i]] = 3;
		res[no[n-1]] = 1;
	}
	else if (n == m+1) {
		vector <pair <int, int> > op = {{a, 1}, {b, 2}, {c, 3}}; sort(op.begin(), op.end());
		a = op[0].F; b = op[1].F; c = op[2].F;
		A = a; B = b;
		dfs(0);
		if (fo == -1) return res;
		for (int i = 0; i < n; i++) res[i] = 3;
		int x = fox, o = fo;
		memset(bio, 0, sizeof bio);
		dfs(x);
		vector <pair <int, int> > v;
		for (auto y : adj[x]) {
			v.push_back({siz[y], y});
		}
		sort(all(v));
		memset(bio, 0, sizeof bio);
		if (o == 0) {
			bio[x] = 1;
			ldfs(v.back().S);
			for (int i = 0; i < b; i++) res[no[i]] = 2;
			no.clear();
			ldfs(x);
			for (int i = 0; i < a; i++) res[no[i]] = 1;
		}
		else {
			bio[x] = 1;
			ldfs(v.back().S);
			for (int i = 0; i < a; i++) res[no[i]] = 1;
			no.clear();
			ldfs(x);
			for (int i = 0; i < b; i++) res[no[i]] = 2;
		}
	}
	return res;
}

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