제출 #1131188

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1131188		AnhPham	사이버랜드 (APIO23_cyberland)	C++20	13 / 100	1946 ms	2162688 KiB

cyberland

#include <bits/stdc++.h>
#ifdef ONLINE_JUDGE
#include "cyberland.h"
#endif

using namespace std;

// #define int     long long
#define sz(v)   (int)(v).size()
#define all(v)  (v).begin(), (v).end()

struct EDGE {
    int x, y, c;
};

void minimize(double &ret, double val) {
    if (ret < -0.5)
        ret = val;
    
    ret = min(ret, val);
}

namespace sub1 {
    bool check_condition(int N, int M, int K, int H, vector <int> &A) {
        return N <= 3 && K <= 30;
    }

    double solve(int N, int M, int K, int H, vector <int> &A, vector <EDGE> &edge) {
        if (H == 0)
            return 0.0;

        vector <vector <double>> cost(N, vector <double> (N, -1));
        for (auto [u, v, c] : edge)
            cost[u][v] = cost[v][u] = c;

        if (N == 2)
            return (cost[0][H] >= 0 ? cost[0][H] : -1);
        
        double ret = (cost[0][H] >= 0 ? cost[0][H] : -1);
        int other = 3 - H;
        if (cost[0][other] >= 0 && cost[other][H] >= 0) {
            if (A[other] == 0)
                minimize(ret, cost[other][H]);
            else if (A[other] == 1)
                minimize(ret, cost[0][other] + cost[other][H]);
            else {
                if (K >= 1)
                    minimize(ret, cost[0][other] / 2.0 + cost[other][H]);
                else
                    minimize(ret, cost[0][other] + cost[other][H]);
            }
        }

        return ret;
    }
}

namespace sub2 {
    bool check_condition(int N, int M, int K, int H, vector <int> &A) {
        return M == N - 1 && K <= 30 && count(all(A), 1) == N;
    }

    vector <vector <pair <int, int>>> adj;
    vector <long long> dp;

    void dfs(int u, int p) {
        for (auto [v, c] : adj[u]) if (v != p) {
            dp[v] = dp[u] + c;
            dfs(v, u);
        }
    }

    double solve(int N, int M, int K, int H, vector <int> &A, vector <EDGE> &edge) {
        adj = vector <vector <pair <int, int>>> (N, vector <pair <int, int>> ());
        for (auto [u, v, c] : edge) {
            adj[u].push_back({ v, c });
            adj[v].push_back({ u, c });
        }
        
        dp = vector <long long> (N, 0ll);
        dfs(0, -1);

        return dp[H];
    }
}

namespace sub3 {
    bool check_condition(int N, int M, int K, int H, vector <int> &A) {
        return M == N - 1 && K <= 30 && *max_element(all(A)) <= 1;
    }

    vector <vector <pair <int, int>>> adj;
    vector <long long> dp;
    queue <int> qu;

    void dfs1(int u, int p, vector <int> &A) {
        for (auto [v, c] : adj[u]) if (v != p) {
            if (A[v] == 0)
                qu.push(v);

            dfs1(v, u, A);
        }
    }

    void dfs2(int u, int p) {
        for (auto [v, c] : adj[u]) if (v != p) {
            dp[v] = min(dp[v], dp[u] + c);
            dfs2(v, u);
        }
    }

    double solve(int N, int M, int K, int H, vector <int> &A, vector <EDGE> &edge) {
        adj = vector <vector <pair <int, int>>> (N, vector <pair <int, int>> ());
        for (auto [u, v, c] : edge) {
            adj[u].push_back({ v, c });
            adj[v].push_back({ u, c });
        }
        
        dfs1(0, -1, A);
        if (A[0] == 0)
            qu.push(0);

        dp = vector <long long> (N, 1e18);
        while (sz(qu)) {
            dp[qu.front()] = 0;
            dfs2(qu.front(), -1);
            qu.pop();
        }

        return dp[H];
    }
}

double solve(int N, int M, int K, int H, vector <int> X, vector <int> Y, vector <int> C, vector <int> A) {
    vector <EDGE> edge;
    for (int i = 0; i < M; ++i)
        edge.push_back({ X[i], Y[i], C[i] });
    
    if (sub1 :: check_condition(N, M, K, H, A))
        return sub1 :: solve(N, M, K, H, A, edge);
    else if (sub2 :: check_condition(N, M, K, H, A))
        return sub2 :: solve(N, M, K, H, A, edge);
    else if (sub3 :: check_condition(N, M, K, H, A))
        return sub3 :: solve(N, M, K, H, A, edge);

    return -1;
}

// main() {
//     int T; cin >> T;
//     while (T--) {
//         int N, M, K, H; cin >> N >> M >> K >> H;

//         vector <int> A(N);
//         for (int &a : A)
//             cin >> a;

//         vector <int> X(M), Y(M), C(M);
//         for (int i = 0; i < M; ++i)
//             cin >> X[i] >> Y[i] >> C[i];

//         cout << (long long)solve(N, M, K, H, X, Y, C, A) << '\n';
//     }
// }

• Subtask 1: 5 / 5
• Subtask 2: 8 / 8
• Subtask 3: 0 / 13
• Subtask 4: 0 / 19
• Subtask 5: 0 / 7
• Subtask 6: 0 / 16
• Subtask 7: 0 / 29
• Subtask 8: 0 / 3