Submission #952134

#	Time	Username	Problem	Language	Result	Execution time	Memory
952134		GrindMachine	Worst Reporter 4 (JOI21_worst_reporter4)	C++17	100 / 100	481 ms	141992 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);
vector<bool> on_cyc(N);

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

map<ll,ll> mp[N];

void dfs2(ll u){
    pll best = {-1,-1};
    trav(v,adj[u]){
        if(on_cyc[v]) conts;
        pll px = {subsiz[v],v};
        amax(best,px);
    }

    ll heavy = best.ss;
    if(heavy == -1){
        auto [h,c] = a[u];
        mp[u][1] = 0;
        mp[u][h+1] = c;
        return;
    }

    dfs2(heavy);
    if(on_cyc[u]){
        mp[u] = mp[heavy];
    }
    else{
        swap(mp[u],mp[heavy]);
    }

    trav(v,adj[u]){
        if(on_cyc[v]) conts;
        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][1] += c;
    mp[u][h] -= c;
    mp[u][h+1] += c;

    auto it = mp[u].find(h);
    vector<ll> rem;

    ll val = mp[u][h], recent_h = h; 
    while(val < 0){
        it--;
        rem.pb(recent_h);
        recent_h = it->first;
        val += it->second;
    }

    mp[u][recent_h] = val;

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

void solve(int test_case)
{
    ll n; cin >> n;
    vector<ll> par(n+5);

    rep1(i,n){
        ll p,h,c; cin >> p >> h >> c;
        adj[p].pb(i);
        par[i] = p;
        a[i] = {h,c};
    }

    vector<ll> outdeg(n+5);
    rep1(i,n) outdeg[i] = sz(adj[i]);

    queue<ll> q;
    rep1(i,n){
        if(outdeg[i] == 0){
            q.push(i);
        }
    }

    while(!q.empty()){
        ll u = q.front();
        q.pop();

        ll p = par[u];
        outdeg[p]--;
        if(outdeg[p] == 0){
            q.push(p);
        }
    }

    rep1(i,n){
        if(outdeg[i]){
            on_cyc[i] = 1;
        }
    }

    vector<bool> vis(n+5);
    ll ans = 0;

    rep1(i,n){
        if(vis[i] or !on_cyc[i]) conts;
        ll j = i;
        map<ll,ll> sum_mp;
        vector<ll> nodes;
        ll tot_sum = 0;

        while(!vis[j]){
            vis[j] = 1;
            nodes.pb(j);
            dfs1(j);
            dfs2(j);
            sum_mp[a[j].ff] += a[j].ss;
            tot_sum += a[j].ss;
            j = par[j];
        }

        map<ll,ll> cost_mp;
        cost_mp[1] += 0;
        ll cost_sum = 0;

        trav(u,nodes){
            cost_mp[a[u].ff] += 0;
            trav(v,adj[u]){
                if(on_cyc[v]) conts;
                for(auto [i,x] : mp[v]){
                    cost_mp[i] += x;
                }
            }
        }

        ll cost = inf2;

        for(auto [h,delta] : cost_mp){
            cost_sum += delta;
            ll curr_cost = cost_sum+tot_sum-sum_mp[h];
            amin(cost,curr_cost);
        }

        ans += cost;
    }

    cout << ans << endl;
}

int main()
{
    fastio;

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

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

    return 0;
}

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