Submission #530646

#	Time	Username	Problem	Language	Result	Execution time	Memory
530646		huangqr	Financial Report (JOI21_financial)	C++14	48 / 100	4064 ms	12780 KiB

Main

#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<ld,ld> pd;
typedef pair<ll,ll> pl;
typedef pair<ll,pl> ppl;
const ll lim=3e5+5,mod=1e9+7;
ll par[lim],a[lim],dp[lim]; //Calced vals can be updated later. In a strict DP sense it would be dp(pos, time) rather than just dp(pos)

ll fs(ll x){
	if(x==par[x])return x;
	else return par[x]=fs(par[x]);
}

void mergeset(ll a,ll b){
	a=fs(a),b=fs(b);
	if(a==b)return;
	par[a]=b;
}

int main(){
	ios_base::sync_with_stdio(0);cin.tie(0);
	ll n,d;
	cin>>n>>d;
	vector<pl>v;
	for(int i=0;i<n;i++){
		cin>>a[i];
		par[i]=i;
		v.emplace_back(a[i],-i);
	}
	sort(v.begin(),v.end());
	for(auto p:v){
		ll i=-p.second;
		dp[i]=max(1ll,dp[i]);
		
		for(int j=max(i-d,0ll);j<i;j++)dp[i]=max(dp[i],dp[j]+1);
		
		ll t = fs(i); //t is the rightmost node to which we can push a dp-val update
		while(t < min(i+d,n-1)){
			mergeset(t, t+1);
			t = fs(t);
		}
		
		for(int j=i+1;j<=t;j++){
			if(dp[j])dp[j]=max(dp[j],dp[i]);
		}
		//cout<<"after "<<i<<" (val "<<p.first<<"):\n";
		//for(int i=0;i<n;i++)cout<<dp[i]<<" ";
		//cout<<"\n";
	}
	
	cout<<dp[n-1];
	
	return 0;
}

• Subtask 1: 14 / 14
• Subtask 2: 14 / 14
• Subtask 3: 20 / 20
• Subtask 4: 0 / 12
• Subtask 5: 0 / 5
• Subtask 6: 0 / 35