IOI 2019 – Split the Attractions Problem and Solution

IOI 2019 – Split the Attractions Problem and Solution

The 31st International Olympiad in Informatics was held in Baku, Azerbaijan in 2019. There were two competition days, with 3 tasks given to the competitors on each day. You can see Split the Attractions Problem and Solution below.

Problem :

ioi2019problem2

Solution :

#include "split.h"
#include <bits/stdc++.h>
#define sz(v) ((int)(v).size())
using namespace std;
using pi = pair<int, int>;
const int MAXN = 100005;

int n;
vector<int> gph[MAXN];
vector<int> tr[MAXN];

namespace report{
	vector<int> gph[MAXN];
	int mark[MAXN], vis[MAXN];
	void dfs(int x, vector<int> &dfn){
		dfn.push_back(x);
		vis[x] = 1;
		for(auto &i : gph[x]){
			if(mark[x] == mark[i] && !vis[i]){
				dfs(i, dfn);
			}
		}
	}
	vector<pi> color;
	vector<int> Do(vector<int> S){
		for(auto &i : S) mark[i] = 1;
		vector<int> ans(n, color[2].second);
		for(int i=0; i<n; i++){
			vector<int> dfn;
			if(mark[i] == 1 && !vis[i]){
				dfs(i, dfn);
				for(int j=0; j<color[0].first; j++){
					ans[dfn[j]] = color[0].second;
				}
			}
			if(mark[i] == 0 && !vis[i]){
				dfs(i, dfn);
				for(int j=0; j<color[1].first; j++){
					ans[dfn[j]] = color[1].second;
				}
			}
		}
		return ans;
	}
}

struct disj{
	int pa[MAXN], sz[MAXN];
	void init(int n){
		iota(pa, pa + n + 1, 0);
		fill(sz, sz + n + 1, 1);
	}
	int find(int x){
		return pa[x] = (pa[x] == x ? x : find(pa[x]));
	}
	int getsz(int x){ return sz[find(x)]; }
	bool uni(int p, int q){
		p = find(p); q = find(q);
		if(p == q) return 0;
		sz[p] += sz[q];
		pa[q] = p; 
		return 1;
	}
}disj;

int sz[MAXN], msz[MAXN];

void dfsc(int x, int p){
	sz[x] = 1; msz[x] = 0;
	for(auto &i : tr[x]){
		if(i != p){
			dfsc(i, x);
			sz[x] += sz[i];
			msz[x] = max(msz[x], sz[i]);
		}
	}
}

int get_center(){
	dfsc(0, 0);
	pi ret(1e9, 1e9);
	for(int i=0; i<n; i++){
		ret = min(ret, pi(max(msz[i], n - sz[i]), i));
	}
	return ret.second;
}

void dfs(int x, int p, vector<int> &dfn){
	dfn.push_back(x);
	for(auto &i : tr[x]){
		if(i != p){
			disj.uni(x, i);
			dfs(i, x, dfn);
		}
	}
}

bool vis[MAXN];

void dfsa(int x, vector<int> &dfn){
	dfn.push_back(x);
	vis[x] = 1;
	for(auto &i : gph[x]){
		if(!vis[i]) dfsa(i, dfn);
	}
}

vector<int> find_split(int _n, int a, int b, int _c, vector<int> p, vector<int> q) {
	n = _n;
	vector<pi> cs = {pi(a, 1), pi(b, 2), pi(_c, 3)};
	sort(cs.begin(), cs.end());
	report::color = cs;
	a = cs[0].first;
	disj.init(n);
	for(int i=0; i<sz(p); i++){
		if(disj.uni(p[i], q[i])){
			tr[p[i]].push_back(q[i]);
			tr[q[i]].push_back(p[i]);
		}
		report::gph[p[i]].push_back(q[i]);
		report::gph[q[i]].push_back(p[i]);
	}
	disj.init(n);
	int c = get_center();
	for(auto &i : tr){
		vector<int> dfn;
		dfs(i, c, dfn);
		if(disj.getsz(i) >= a){
			return report::Do(dfn);
		}
	}
	for(int i=0; i<sz(p); i++){
		if(p[i] == c || q[i] == c) continue;
		int l = disj.find(p[i]);
		int r = disj.find(q[i]);
		if(l != r){
			gph[l].push_back(r);
			gph[r].push_back(l);
		}
	}
	for(int i=0; i<n; i++){
		if(i != c && disj.find(i) == i && !vis[i]){
			vector<int> dfn;
			dfsa(i, dfn);
			int sum = 0;
			set<int> pot;
			for(auto &i : dfn){
				sum += disj.getsz(i);
				pot.insert(i);
				if(sum >= a) break;
			}
			if(sum >= a){
				vector<int> ans;
				for(int i=0; i<n; i++){
					if(pot.find(disj.find(i)) != pot.end()){
						ans.push_back(i);
					}
				}
				return report::Do(ans);
			}
		}
	}
	return vector<int>(n, 0);
}

Explanation :

split

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the above hole problem statement is given by hackerrank.com but the solution is generated by the SLTECHACADEMY authority if any of the query regarding this post or website fill the following contact form thank you.

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