제출 #1032432

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1032432		Turkhuu	봉쇄 시간 (IOI23_closing)	C++17	0 / 100	949 ms	2097152 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, ll K, vector<int> U, vector<int> V, vector<int> W) {
    vector<vector<array<int, 2>>> 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]});
    }
    auto bfs = [&](int s) {
        vector<ll> dis(N);
        vector<int> from(N, -1);
        from[s] = s;
        queue<int> q;
        q.push(s);
        while (!q.empty()) {
            int x = q.front();
            q.pop();
            for (auto [y, z] : adj[x]) {
                if (from[y] == -1) {
                    from[y] = x;
                    dis[y] = dis[x] + z;
                    q.push(y);
                }
            }
        }
        return make_pair(dis, from);
    };
    auto [dis_x, from_x] = bfs(X);
    auto [dis_y, from_y] = bfs(Y);
    vector<ll> one(N), two(N);
    for (int i = 0; i < N; i++) {
        tie(one[i], two[i]) = minmax(dis_x[i], dis_y[i]);
    }
    int ans = 0; {
        auto a(one);
        sort(a.begin(), a.end());
        ll sum = 0;
        while (ans < N && sum + a[ans] <= K) sum += a[ans++];
    }
    vector<int> path{X};
    while (path.back() != Y) path.push_back(from_y[path.back()]);
    int M = path.size();
    vector<int> on_path(N);
    for (auto x : path) on_path[x] = 1;
    vector dp(N, vector(3, vector<ll>(2 * N + 1, 1e18)));
    function<void(int, int)> dfs = [&](int x, int p) {
        dp[x][0][0] = 0;
        dp[x][1][1] = one[x];
        dp[x][2][2] = two[x];
        for (auto [y, z] : adj[x]) {
            if (y == p || on_path[y]) continue;
            dfs(y, x);
            for (int i = 1; i < 2 * N; i++) {
                for (int j = 2 * N; j > i; j--) {
                    dp[x][1][j] = min(dp[x][1][j], dp[y][1][i] + dp[x][1][j - i]);
                    dp[x][2][j] = min(dp[x][2][j], min(dp[y][1][i], dp[y][2][i]) + dp[x][2][j - i]);
                }
            }
        }
    };
    for (auto x : path) dfs(x, -1);
    vector f(M + 1, vector(3, vector<ll>(2 * N + 1, 1e18)));
    f[0][0][0] = 0;
    for (int i = 1; i <= M; i++) {
        int x = path[i - 1];
        for (int j = 0; j <= 2 * N; j++) {
            for (int k = j; k <= 2 * N; k++) {
                f[i][0][k] = min(f[i][0][k], f[i - 1][0][j] + dp[x][1][k - j]);
                f[i][2][k] = min(f[i][2][k], min(f[i - 1][0][j], f[i - 1][2][j]) + dp[x][2][k - j]);
                f[i][2][k] = min(f[i][2][k], min(f[i - 1][1][j], f[i - 1][2][j]) + dp[x][1][k - j]);
            }
        }
    }
    for (int i = 2 * N; i >= 0; i--) {
        if (f[M][1][i] <= K || f[M][2][i] <= K) {
            ans = max(ans, i);
        }
    }
    return ans;
}

• 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