Submission #481208

#	Time	Username	Problem	Language	Result	Execution time	Memory
481208		CSQ31	Papričice (COCI20_papricice)	C++17	50 / 110	102 ms	27108 KiB

papricice

#pragma GCC optimize("Ofast") 
#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define fi first
#define se second
#define sz(a) (int)(a.size())
#define all(a) a.begin(),a.end()
#define lb lower_bound
#define ub upper_bound
#define owo ios_base::sync_with_stdio(0);cin.tie(0);
#define MOD (ll)(998244353)
#define INF (ll)(1e18)
#define debug(...) fprintf(stderr, __VA_ARGS__),fflush(stderr)
#define time__(d) for(long blockTime = 0; (blockTime == 0 ? (blockTime=clock()) != 0 : false);\
debug("%s time : %.4fs\n", d, (double)(clock() - blockTime) / CLOCKS_PER_SEC))
typedef long long int ll;
typedef long double ld;
typedef pair<ll,ll> PII;
typedef pair<int,int> pii;
typedef vector<vector<int>> vii;
typedef vector<vector<ll>> VII;
ll gcd(ll a,ll b){if(!b)return a;else return gcd(b,a%b);}
const int MAXN = 2e5; 
int ans = 1e9,n;
vii adj(MAXN);
int sub[MAXN];
multiset<int>a,b;
void dfs0(int v,int u){
	sub[v] = 1;
	for(int x:adj[v]){
		if(x==u)continue;
		dfs0(x,v);
		sub[v]+=sub[x];
	}
}
void dfs(int v,int u){
	//sub[i] ,sub[j] and n-sub[i]-sub[j]
	//we want sub[j] and n-sub[i]-sub[j] to be as close as possible
	//sub[j] = n-sub[i]-sub[j]
	//sub[j] = n-sub[i]/2;
	auto it = a.lb((n-sub[v])/2);
	if(it != a.end())ans = min(ans,max({sub[v],*it,n-sub[v]-*it}) - min({sub[v],*it,n-sub[v]-*it}));
	if(it != a.begin()){
		it--;
		ans = min(ans,max({sub[v],*it,n-sub[v]-*it}) - min({sub[v],*it,n-sub[v]-*it}));
	}
	
	//ancestor
	//sub[i] , n-sub[j] , sub[j] - sub[i];
	//n - sub[j] = sub[j] - sub[i]
	//sub[j] = (n+sub[i])/2
	it = b.lb((n+sub[v])/2);
	if(it != b.end())ans = min(ans,max({sub[v],*it-sub[v],n-*it}) - min({sub[v],*it-sub[v],n-*it}));
	if(it != b.begin()){
	    it--;
	    ans = min(ans,max({sub[v],*it-sub[v],n-*it}) - min({sub[v],*it-sub[v],n-*it}));
    }
	
	b.insert(sub[v]);
	for(int x:adj[v]){
		if(x==u)continue;
		dfs(x,v);
	}
	b.erase(b.find(sub[v]));
	a.insert(sub[v]);
	
}
int main()
{
	owo
	cin>>n;
	for(int i=0;i<n-1;i++){
		int v,u;
		cin>>v>>u;
		adj[v].pb(u);
		adj[u].pb(v);
	}
	dfs0(1,0);
	dfs(1,0);
	cout<<ans;
	
}

• Subtask 1: 15 / 15
• Subtask 2: 35 / 35
• Subtask 3: 0 / 60