Submission #425216

#	Time	Username	Problem	Language	Result	Execution time	Memory
425216		alishahali1382	Split the Attractions (IOI19_split)	C++14	100 / 100	166 ms	16240 KiB

split

#include "split.h"
#include<bits/stdc++.h>
#pragma GCC optimize("O2")
using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
#define debug(x) {cerr<<#x<<"="<<x<<"\n";}
#define debug2(x, y) {cerr<<"{"<<#x<<", "<<#y<<"}={"<<x<<", "<<y<<"}\n";}
#define debugp(p) {cerr<<#p<<"={"<<p.first<<", "<<p.second<<"}\n";}
#define debugv(abcd) {cerr<<#abcd<<": ";for (auto dcba:abcd) cerr<<dcba<<", ";cerr<<"\n";}
#define pb push_back
#define SZ(x) ((int)x.size())
#define all(x) x.begin(), x.end()

const int inf=1000001000; // 1e9
const ll INF=10000000010000000; // 1e16
const int mod=1000000007;
const int MAXN=100010;

vector<pair<int, char>> perm;
int n, m, k, a, b, c, u, v, t, x, y, solved;
int mark[MAXN], sz[MAXN], par[MAXN];
vector<int> G[MAXN];
vector<int> out;

int dfs1(int node){
	mark[node]=1;
	for (int v:G[node]) if (!mark[v]){
		par[v]=node;
		sz[node]+=dfs1(v);
	}
	return ++sz[node];
}
void dfs2(int node){
	mark[node]=2;
	for (int v:G[node]) if (mark[v]!=2) dfs2(v);
}
void dfs3(int node, vector<int> &topol){
	mark[node]=3;
	topol.pb(node);
	for (int v:G[node]) if (mark[v]!=3 && par[v]==node) dfs3(v, topol);
}

vector<int> find_split(int _n, int _a, int _b, int _c, vector<int> _p, vector<int> _q){
	n=_n;
	out.resize(n, 0);
	a=_a;
	b=_b;
	c=_c;
	for (int i=0; i<SZ(_p); i++){
		int u=_p[i], v=_q[i];
		G[u].pb(v);
		G[v].pb(u);
	}
	perm.pb({a, 'a'});
	perm.pb({b, 'b'});
	perm.pb({c, 'c'});
	sort(all(perm));
	a=perm[0].first;
	b=perm[1].first;
	c=perm[2].first;

	vector<int> X, Y;
	dfs1(0);
	int v=0;
	for (int i=1; i<n; i++) if (sz[i]>=a && sz[i]<sz[v]) v=i;
	if (!v) return out; // -1
	mark[v]=2;
	dfs2(0);
	mark[v]=3;
	X.pb(v);
	dfs3(0, Y);
	int x=sz[v], y=n-x;
	for (int u:G[v]) if (par[u]==v){
		if (mark[u]!=2 || x-a<sz[u]) dfs3(u, X);
		else{
			x-=sz[u];
			y+=sz[u];
			dfs3(u, Y);
		}
	}
	if (y>=b){
		for (int i=0; i<a; i++) out[X[i]]=1;
		for (int i=0; i<b; i++) out[Y[i]]=2;
		solved=1;
	}
	else if (y>=a){
		for (int i=0; i<a; i++) out[Y[i]]=1;
		for (int i=0; i<b; i++) out[X[i]]=2;
		solved=1;
	}

	if (!solved) return out;
	for (int i=0; i<n; i++){
		if (!out[i]) out[i]=3;
		out[i]=perm[out[i]-1].second-'a'+1;
	}
	return out;
}

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