Submission #1260751

#	Time	Username	Problem	Language	Result	Execution time	Memory
1260751		LittleCube	봉쇄 시간 (IOI23_closing)	C++17	43 / 100	144 ms	38972 KiB

closing

#include "closing.h"

#include <vector>
#include <utility>
#include <algorithm>
#include <limits>
#include <queue>
using namespace std;

int max_score(int N, int X, int Y, long long K,
              std::vector<int> U, std::vector<int> V, std::vector<int> W)
{
    int ans = 0;

    vector<vector<pair<int, int>>> adj(N);
    vector<long long> distX(N), distY(N);
    for (int i = 0; i < N - 1; i++)
    {
        adj[U[i]].emplace_back(V[i], W[i]);
        adj[V[i]].emplace_back(U[i], W[i]);
    }

    {
        auto dfs = [&](auto self, int u, vector<long long> &d, vector<bool> &vis) -> void
        {
            vis[u] = true;
            for (auto [v, w] : adj[u])
                if (!vis[v])
                {
                    d[v] = d[u] + w;
                    self(self, v, d, vis);
                }
        };
        vector<bool> vis(N, false);
        distX[X] = 0;
        dfs(dfs, X, distX, vis);
        distY[Y] = 0;
        fill(vis.begin(), vis.end(), false);
        dfs(dfs, Y, distY, vis);
    }

    // Case 1: no intersections
    {
        long long remain = K;
        int score = 0;

        vector<long long> single;
        single.insert(single.end(), distX.begin(), distX.end());
        single.insert(single.end(), distY.begin(), distY.end());
        sort(single.begin(), single.end());

        for (auto cost : single)
            if (remain >= cost)
                score++, remain -= cost;
        ans = max(ans, score);
    }

    // Case 2: intersecting
    {
        long long remain = K;
        int root = 0;
        int score = 0;
        {
            long long min_dist = numeric_limits<long long>::max();
            for (int i = 0; i < N; i++)
                if (min_dist > max(distX[i], distY[i]))
                    min_dist = max(distX[i], distY[i]), root = i;

            vector<bool> take_two(N, false);
            vector<pair<long long, int>> one, two;
            priority_queue<long long> used_two;

            for (int i = 0; i < N; i++)
                if (i == root)
                    score += 2, remain -= min_dist;
                else
                {
                    long long first = min(distX[i], distY[i]);
                    long long second = max(distX[i], distY[i]) - first;

                    if (distX[i] + distY[i] == distX[root] + distY[root])
                        score++, remain -= first, one.emplace_back(second, i);
                    else
                    {
                        if (second >= first)
                            one.emplace_back(first, i), one.emplace_back(second, i);
                        else
                            one.emplace_back(first, i), two.emplace_back(first + second, i);
                    }
                }

            sort(one.begin(), one.end(), greater<>());
            sort(two.begin(), two.end(), greater<>());

            while (remain >= 0)
            {
                for (size_t t = 1; t <= 2; t++)
                    while (one.size() >= t && take_two[(one.end() - t)->second])
                        one.erase(one.end() - t);

                // Case 1: as-is
                ans = max(ans, score);

                // Case 1: take one
                if (!one.empty() && remain - one.back().first >= 0)
                    ans = max(ans, score + 1);
                // Case 2: two-bundle decomposing and it is one of the previous added bundle
                if (!two.empty())
                {
                    auto [cost, i] = two.back();
                    long long revert_earn = max(distX[i], distY[i]) - min(distX[i], distY[i]);
                    if (!used_two.empty())
                        revert_earn = max(revert_earn, used_two.top());
                    if (remain - cost + revert_earn >= 0)
                        ans = max(ans, score + 1);
                }

                if (!two.empty())
                {
                    auto [cost, i] = two.back();
                    if (one.size() < 2 || one.back().first + (one.rbegin() + 1)->first >= cost)
                    {
                        score += 2;
                        take_two[i] = true;
                        remain -= cost;
                        two.pop_back();
                        used_two.emplace(max(distX[i], distY[i]) - min(distX[i], distY[i]));
                        continue;
                    }
                }
                auto [cost, i] = one.back();
                score++;
                remain -= cost;
                one.pop_back();
            }
        }
    }

    return ans;
}

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