Submission #1213122

#	Time	Username	Problem	Language	Result	Execution time	Memory
1213122		trimkus	Closing Time (IOI23_closing)	C++20	0 / 100	1016 ms	2162688 KiB

closing

#include "closing.h"
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int max_score(int N, int X, int Y, long long K,
              std::vector<int> U, std::vector<int> V, std::vector<int> _W)
{
    vector<vector<pair<int, int>>> adj(N);
    for (int i = 0; i < N - 1; ++i) {
        adj[U[i]].push_back({V[i], _W[i]});
        adj[V[i]].push_back({U[i], _W[i]});
    }
    vector<ll> C(N);
    vector<vector<bool>> vis(N, vector<bool>(2));
    priority_queue<array<ll, 4>> pq;
    pq.push({0, X, 0, 0});
    pq.push({0, Y, 0, 1});
    ll tot = 0;
    int res = 0;
    while (pq.size()) {
        ll need = -pq.top()[0];
        int v = pq.top()[1];
        ll sum = pq.top()[2];
        int p = pq.top()[3];
        pq.pop();
        if (sum - C[v] != need) {
            pq.push({-max(0LL, sum - C[v]), v, sum, p});
            continue;
        }
        if (need + tot > K) continue;
        tot += need;
        res += 1;
        C[v] += need;
        vis[v][p] = 1;
        for (auto& [u, w] : adj[v]) {
            if (vis[u][p]) continue;
            pq.push({-max(0LL, sum + w - C[u]), u, sum + w, p});
        }
    }
    cerr << "Now dp:\n";
    vector<vector<ll>> dist(2, vector<ll>(N));
    auto dfs = [&](auto& dfs, int v, int p, int i) -> void {
        for (auto& [u, w] : adj[v]) {
            if (u == p) continue;
            dist[i][u] = dist[i][v] + w;
            dfs(dfs, u, v, i);
        }
    };
    
    dfs(dfs, X, -1, 0);
    dfs(dfs, Y, -1, 1);
    const ll INF = 1e18;
    vector<vector<ll>> dp(N + 1, vector<ll>(2 * N + 5, INF));
    dp[0][0] = 0;
    for (int i = 0; i < N; ++i) {
        for (int j = 0; j <= 2 * N; ++j) {
            // if (dist[0][i] + dist[1][i] == dist[0][Y]) {
            //     dp[i + 1][j + 1] = min(dp[i + 1][j + 1], dp[i][j] + min(dist[0][i], dist[1][i]));
            //     dp[i + 1][j + 2] = min(dp[i + 1][j + 2], dp[i][j] + max(dist[0][i], dist[1][i]));
            //     continue;
            // }
            dp[i + 1][j] = min(dp[i + 1][j], dp[i][j]);
            dp[i + 1][j + 1] = min(dp[i + 1][j + 1], dp[i][j] + min(dist[0][i], dist[1][i]));
            dp[i + 1][j + 2] = min(dp[i + 1][j + 2], dp[i][j] + max(dist[0][i], dist[1][i]));
        }
    }
    for (int i = 0; i <= 2 * N; ++i) {
        if (dp[N][i] <= K) {
            res = max(res, i);
        }
    }
    return res;
}

• Subtask 0: 0 / 0
• Subtask 1: 0 / 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