Submission #952045

#	Time	Username	Problem	Language	Result	Execution time	Memory
952045		GrindMachine	Worst Reporter 4 (JOI21_worst_reporter4)	C++17	0 / 100	9 ms	19548 KiB

worst_reporter2

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) (int)a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x,y) ((x+y-1)/(y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i,n) for(int i = 0; i < n; ++i)
#define rep1(i,n) for(int i = 1; i <= n; ++i)
#define rev(i,s,e) for(int i = s; i >= e; --i)
#define trav(i,a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a,b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a,b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif

/*

refs:
https://codeforces.com/blog/entry/88748?#comment-774017 (thread)
got the key idea, but didnt know how to implement in a clean way

implementation idea:
represent the subtree cost @u as a staircase function using a map
can be merged efficiently using small to large

*/

const int MOD = 1e9 + 7;
const int N = 2e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

vector<ll> adj[N];
vector<pll> a(N);
vector<ll> subsiz(N);

void dfs1(ll u){
    subsiz[u] = 1;
    trav(v,adj[u]){
        dfs1(v);
        subsiz[u] += subsiz[v];
    }
}

map<ll,ll> mp[N];

void dfs2(ll u){
    if(adj[u].empty()){
        auto [h,c] = a[u];
        mp[u][1] = 0;
        mp[u][h+1] = c;
        return;
    }

    pll best = {-1,-1};
    trav(v,adj[u]){
        pll px = {subsiz[v],v};
        amax(best,px);
    }

    ll heavy = best.ss;
    dfs2(heavy);
    swap(mp[u],mp[heavy]);

    trav(v,adj[u]){
        if(v == heavy) conts;
        dfs2(v);
        for(auto [i,x] : mp[v]){
            mp[u][i] += x;
        }
        mp[v].clear();
    }

    auto [h,c] = a[u];
    mp[u].begin()->second += c;
    mp[u][h] -= c;
    mp[u][h+1] += c;
    
    auto it = mp[u].find(h);
    vector<ll> rem;

    while(it != mp[u].begin() and mp[u][h] < 0){
        it--;
        mp[u][h] += it->second;
        rem.pb(it->first);
    }

    trav(x,rem){
        mp[u].erase(x);
    }

    if(mp[u].begin()->first != 1){
        auto [x,y] = *mp[u].begin();
        mp[u][1] = y;
        mp[u].erase(x);        
    }
}

void solve(int test_case)
{
    ll n; cin >> n;
    rep1(i,n){
        ll p,h,c; cin >> p >> h >> c;
        if(p != i){
            adj[p].pb(i);
        }
        a[i] = {h,c};
    }

    dfs1(1);
    dfs2(1);

    ll ans = mp[1].begin()->second;
    cout << ans << endl;
}

int main()
{
    fastio;

    int t = 1;
    // cin >> t;

    rep1(i, t) {
        solve(i);
    }

    return 0;
}

• Subtask 1: 0 / 14
• Subtask 2: 0 / 65
• Subtask 3: 0 / 21