Submission #295091

#	Time	Username	Problem	Language	Result	Execution time	Memory
295091		AKaan37	Split the Attractions (IOI19_split)	C++17	0 / 100	675 ms	1048580 KiB

split

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

using namespace std;

typedef long long lo;
typedef pair< lo,lo > PII;

#define fi first
#define se second
#define mp make_pair
#define endl "\n"
#define pb push_back
#define fio() ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define FOR for(int i=1;i<=n;i++)
#define mid ((start+end)/2)
#define ort ((bas+son)/2)

const lo inf = 2000000000;
const lo KOK = 100000;
const lo LOG = 30;
const lo li = 500005;
const lo mod = 1000000007;

int n,m,b[li],a[li],k,flag,t,ind,mn=inf,mx,mx1,par[li],vis[li],say,gel,sub[li];
int cev;
string s;
vector<int> v[li];

inline void dfs(int node,int ata){
	if(vis[node])return;
	sub[node]=1;
	for(int i=0;i<(int)v[node].size();i++){
		int go=v[node][i];
		if(go==ata)continue;
		if(vis[go])continue;
		dfs(go,node);
		sub[node]+=sub[go];
	}
	par[node]=ata;
	if(sub[node]>=mx && sub[node]<mn){ind=node;mn=sub[node];}
}

inline void dfs1(int node,int ata,int aa,int bb,int cc){
	if(vis[node])return;
	if(mx==aa && (b[1]==0 || b[1]==gel)){vis[node]=1;b[1]=gel;say++;}
	else if(mx==bb && (b[2]==0 || b[2]==gel)){vis[node]=2;b[2]=gel;say++;}
	else if(mx==cc && (b[3]==0 || b[3]==gel)){vis[node]=3;b[3]=gel;say++;}
	if(say>=mx)return ;
	for(int i=0;i<(int)v[node].size();i++){
		int go=v[node][i];
		if(go==ata)continue;
		dfs1(go,node,aa,bb,cc);
	}
}

vector<int> find_split(int n, int aa, int bb, int cc, vector<int> p, vector<int> q) {
	vector<int> res;
	a[1]=aa;
	a[2]=bb;
	a[3]=cc;
	sort(a+1,a+4);
	cerr<<aa<<bb<<cc;
	mx=a[2];
	mx1=a[1];
	for(int i=0;i<(int)p.size();i++){
		v[p[i]].pb(q[i]);
		v[q[i]].pb(p[i]);
	}
	gel=1;
	ind=-1;
	dfs(0,-1);
	if(ind==-1){
		FOR res.pb(0);
		return res;
	}
	dfs1(ind,par[ind],aa,bb,cc);
	mx=mx1;
	gel=2;
	say=0;
	mn=inf;
	ind=-1;
	dfs(0,-1);
	if(ind==-1){
		FOR res.pb(0);
		return res;
	}
	dfs1(ind,par[ind],aa,bb,cc);
	for(int i=0;i<n;i++){
		if(vis[i])res.pb(vis[i]);
		else{
			if(b[1]==0)res.pb(1);
			if(b[2]==0)res.pb(2);
			if(b[3]==0)res.pb(3);
		}
	}
	return res;
}

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