제출 #1357982

#제출 시각아이디문제언어결과실행 시간메모리
1357982maya_sClosing Time (IOI23_closing)C++20
43 / 100
162 ms48556 KiB
#include "closing.h"
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pll;

void calc_dist(ll n, ll p, vector<ll> &dist, vector<vector<pll>> &g){
    for(auto[i, d]: g[n]) if(i != p){
        dist[i] = dist[n] + d;
        calc_dist(i, n, dist, g);
    }
}

ll max_without_overlap(vector<vector<ll>> &dist, ll n, ll k){
    ll ans = 0, credit = k;
    priority_queue<pll, vector<pll>, greater<pll>> pq;
    for(ll i = 0; i < n; i++) pq.push({dist[0][i], i}), pq.push({dist[1][i], i});
    vector<bool> vis(n);
    while(pq.size()){
        auto[d, node] = pq.top(); pq.pop();
        if(vis[node]) continue;
        vis[node] = 1;
        if(credit - d < 0) continue;
        credit -= d;
        ans++;
    }
    return ans;
}

int max_score(int N, int X, int Y, long long k, vector<int> u, vector<int> v, vector<int> w)
{
    ll n = N, x = X, y = Y;
    vector<vector<pll>> g(n);
    for(ll i = 0; i < n-1; i++) g[u[i]].push_back({v[i], w[i]}), g[v[i]].push_back({u[i], w[i]});
    vector<vector<ll>> dist(2, vector<ll>(n)), pref(2, vector<ll>(n+1));
    calc_dist(x, -1, dist[0], g);
    calc_dist(y, -1, dist[1], g);
    ll ans = max_without_overlap(dist, n, k);
    for(ll i = 0; i <= x; i++) {
        multiset<ll, greater<ll>> ms;
        ll s = 0, cnt = 0;
        for(ll j = i; j < y; j++){
            cnt++;
            s += min(dist[0][j], dist[1][j]);
            s += max(dist[0][j], dist[1][j]) - min(dist[0][j], dist[1][j]); 
            cnt++;
            ms.insert(max(dist[0][j], dist[1][j]) - min(dist[0][j], dist[1][j]));
            while(ms.size()){
                if(s <= k) break;
                ll d = *ms.begin(); ms.erase(ms.begin());
                s -= d; cnt--;
            }
            if(s > k) break;
        }
        for(ll j = y; s <= k && j < n; j++) {
            cnt++;
            s += min(dist[0][j], dist[1][j]);
            s += max(dist[0][j], dist[1][j]) - min(dist[0][j], dist[1][j]); 
            cnt++;
            ms.insert(max(dist[0][j], dist[1][j]) - min(dist[0][j], dist[1][j]));
            while(ms.size()){
                if(s <= k) break;
                ll d = *ms.begin(); ms.erase(ms.begin());
                s -= d; cnt--;
            }
            if(s > k) break;
            ans = max(ans, cnt);
        }
    }
    return ans;
}
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