Submission #1179859

#	Time	Username	Problem	Language	Result	Execution time	Memory
1179859		ByeWorld	Magic Tree (CEOI19_magictree)	C++20	0 / 100	119 ms	25812 KiB

magictree

#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#define int long long
#define ll long long
#define pb push_back
#define fi first
#define se second
#define lf (id<<1)
#define rg ((id<<1)|1)
#define md ((l+r)>>1)
#define ld long double
using namespace std;
typedef pair<int,int> pii;
typedef pair<int,pii> ipii;
const int MAXN = 1e5+10;
const int MAXA = 1e9;
const int INF = 2e9+100;
const int SQRT = 500;
const int LOG = 19;
const int MOD = 998244353;
void chmn(auto &a, auto b){ a = min(a, b); }
void chmx(auto &a, auto b){ a = max(a, b); }
int sum(int a, int b){ a %= MOD; b %= MOD; return (a+b)%MOD; }
void chsum(int &a, int b){ a %= MOD; b %= MOD; a = (a+b)%MOD; }
void chsub(int &a, int b){ a %= MOD; b %= MOD; a = (a+MOD-b)%MOD; }
int mul(int a, int b){ a %= MOD; b %= MOD; return a*b%MOD;}
void chmul(int &a, int b){ a = a*b%MOD; }
int expo(int a, int b){
	if(b==0) return 1;
	int te = expo(a, b/2); te = mul(te, te);
	return (b%2 ? mul(te, a) : te);
}

int n, m, k, par[MAXN], d[MAXN], w[MAXN];
vector<int> adj[MAXN];
int dp[MAXN][23];

void bd(int nw){
	for(auto nx : adj[nw]) bd(nx);

	for(int i=1; i<=20; i++){
		dp[nw][i] = dp[nw][i-1];
		int no = 0;
		for(auto nx : adj[nw])
			no += dp[nx][i];
		chmx(dp[nw][i], no);

		if(i==d[nw]){
			int take = 0;
			for(auto nx : adj[nw])
				take += dp[nx][i-1];
			chmx(dp[nw][i], w[nw]+take);
		}
		// if(i<=4) 
		// 	cout << nw << ' ' << i << ' ' << dp[nw][i] << " x\n";
	}
}
signed main(){
	// ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>>n>>m>>k;
	for(int i=2; i<=n; i++){
		cin>>par[i]; adj[par[i]].pb(i);
	}
	vector <int> cc; cc.pb(-1);
	for(int i=1; i<=m; i++){
		int id, day, wei; cin>>id>>day>>wei;
		// vec.pb({day, id});
		d[id] = day; w[id] = wei;
		cc.pb(day);
	}
	sort(cc.begin(), cc.end());
	cc.resize(unique(cc.begin(), cc.end())-cc.begin());
	for(int i=1; i<=n; i++)
		d[i] = lower_bound(cc.begin(), cc.end(), d[i])-cc.begin();

	bd(1);
	cout << dp[1][20] << '\n';
}

• Subtask 1: 0 / 6
• Subtask 2: 0 / 3
• Subtask 3: 0 / 11
• Subtask 4: 0 / 12
• Subtask 5: 0 / 16
• Subtask 6: 0 / 13
• Subtask 7: 0 / 22
• Subtask 8: 0 / 17