Submission #1072066

#	Time	Username	Problem	Language	Result	Execution time	Memory
1072066		mindiyak	Closing Time (IOI23_closing)	C++17	8 / 100	156 ms	22100 KiB

closing

#include "closing.h"
#pragma GCC optimize("O1,O2,O3,Ofast,unroll-loops")
#include <bits/stdc++.h>
#include <string>
#include <iostream>
#include <cmath>
#include <numeric>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pi;
typedef pair<int, int> pl;
typedef pair<ld, ld> pd;
typedef vector<int> vi;
typedef vector<bool> vb;
typedef vector<vector<int>> vvi;
typedef vector<vector<ll>> vvl;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
#define FOR(i, a, b) for (int i = a; i < (b); i++)
#define F0R(i, a) for (int i = 0; i < (a); i++)
#define FORd(i, a, b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i, a) for (int i = (a)-1; i >= 0; i--)
#define trav(a, x) for (auto &a : x)
#define uid(a, b) uniform_int_distribution<int>(a, b)(rng)
#define len(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define F first
#define nl endl
#define S second
#define lb lower_bound
#define ub upper_bound
#define aint(x) x.begin(), x.end()
#define raint(x) x.rbegin(), x.rend()
#define ins insert
const int MOD = 1000000007;

int max_score(int N, int X, int Y, ll K,vi U, vi V, vi W)
{
    vector<vpi> paths(2e5+5);

    FOR(i,0,N-1){
        paths[U[i]].pb({V[i],W[i]});
        paths[V[i]].pb({U[i],W[i]});
    }

    priority_queue<pair<ll,pi>> arr;
    vl  closing_times(N,0);

    int ans = 0;
    arr.push({-0,{X,-1}});
    arr.push({-0,{Y,-1}});

    while(!arr.empty()){
        int node = arr.top().S.F;
        int prev = arr.top().S.S;
        ll diff = -arr.top().F;
        arr.pop();

        cerr << K << " " << diff << " " << node << endl;

        if(diff > K)continue;

        K-=diff;
        ans += 1;
        closing_times[node] += diff;

        for(pi edges:paths[node]){
            if(edges.F == prev)continue;
            arr.push({-(max(closing_times[node]+edges.S-closing_times[edges.F],0LL)),{edges.F,node}});
        }
    }

    return ans;
}

• Subtask 0: 0 / 0
• Subtask 1: 8 / 8
• Subtask 2: 0 / 9
• Subtask 3: 0 / 12
• Subtask 4: 0 / 14
• Subtask 5: 0 / 9
• Subtask 6: 0 / 11
• Subtask 7: 0 / 10
• Subtask 8: 0 / 10
• Subtask 9: 0 / 17