제출 #744518

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
744518		t6twotwo	Road Closures (APIO21_roads)	C++17	0 / 100	2085 ms	28376 KiB

roads

#include "roads.h"
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
constexpr ll inf = 1E18;
vector<ll> minimum_closure_costs(int N, vector<int> U, vector<int> V, vector<int> W) {
    if (N <= 2000) {
        vector<vector<pair<int, int>>> adj(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]);
        }
        vector dp(N, vector<ll>(N, inf));
        auto dfs = [&](auto dfs, int x, int p) -> void {
            dp[x][0] = 0;
            vector<pair<int, int>> ch;
            for (auto [y, z] : adj[x]) {
                if (y == p) {
                    continue;
                }
                dfs(dfs, y, x);
                dp[x][0] += dp[y][0] + z;
                ch.emplace_back(y, z);
            }
            for (int i = 1; i < N; i++) {
                vector<ll> a(i + 1, inf);
                a[0] = 0;
                for (auto [y, z] : ch) {
                    auto b = a;
                    a.assign(i + 1, inf);
                    for (int j = 0; j <= i; j++) {
                        a[j] = min(a[j], b[j] + dp[y][i] + z);
                        if (j + 1 <= i) {
                            a[j + 1] = min(a[j + 1], b[j] + dp[y][i - 1]);
                        }
                    }
                }
                dp[x][i] = *min_element(a.begin(), a.end());
            }
        };
        dfs(dfs, 0, -1);
        return dp[0];
    }
    if (U == vector<int>(N - 1)) {
        vector<ll> ans(N);
        sort(W.begin(),  W.end());
        for (int i = 0; i < N - 1; i++) {
            ans[N - i - 2] = ans[N - i - 1] + W[i];
        }
        return ans;
    }
    ll dp0 = 0, dp1 = 0;
    for (int i = 0; i < N; i++) {
        ll pd0 = dp0;
        dp0 = max(dp0, dp1);
        dp1 = W[i] + pd0;
    }
    vector<ll> ans(N);
    ans[0] = accumulate(W.begin(), W.end(), 0LL);
    ans[1] = ans[0] - max(dp0, dp1);
    return ans;
}

• Subtask 1: 0 / 5
• Subtask 2: 0 / 7
• Subtask 3: 0 / 14
• Subtask 4: 0 / 10
• Subtask 5: 0 / 17
• Subtask 6: 0 / 25
• Subtask 7: 0 / 22