제출 #955400

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
955400		ProtonDecay314	사이버랜드 (APIO23_cyberland)	C++17	15 / 100	3048 ms	527692 KiB

cyberland

#include "cyberland.h"

#include <vector>
#include <bits/stdc++.h>
using namespace std;

struct node {
    int i;
    int k;
    
    node(int a_i, int a_k): i(a_i), k(a_k) {};

    int get_ind() {
        return (i << 5) | k;
    }

    bool operator<(const node& other) const {
        return true;
    }
};

double solve(int N, int M, int K, int H, std::vector<int> x, std::vector<int> y, std::vector<int> c, std::vector<int> arr) {
    vector<vector<pair<int, int>>> adj;
    for(int i = 0; i < N; i++) {
        vector<pair<int, int>> adj_row;
        adj.push_back(adj_row);
    }

    for(int i = 0; i < M; i++) {
        int curx = x[i], cury = y[i], curc = c[i];

        adj[curx].push_back({cury, curc});
        adj[cury].push_back({curx, curc});
    }

    /*
    Initial BFS to find the reachable nodes
    */

    vector<bool> found(N, false);
    queue<int> bfs_q;
    bfs_q.push(0);

    while(!bfs_q.empty()) {
        int curq = bfs_q.front();
        bfs_q.pop();

        if(found[curq]) continue;
        found[curq] = true;

        for(auto [j, w] : adj[curq]) {
            bfs_q.push(j);
        }
    }

    /*
    Idea: Dijkstra's algorithm
    priority_queue<{total_time, node}>
    */
    double ans = numeric_limits<double>::max();

    priority_queue<pair<double, node>, vector<pair<double, node>>, greater<pair<double, node>>> q;

    for(int k = 0; k <= K; k++) {
        q.push({0.0, {0, k}});
        for(int i = 1; i < N; i++) {
            if(arr[i] == 0 && found[i]) {
                q.push({0.0, {i, k}});
            }
        }
    }

    vector<bool> vis((N << 5) | K, false);

    while(! q.empty()) {
        auto [cur_time, cur_node] = q.top();

        q.pop();

        if(vis[cur_node.get_ind()]) continue;
        vis[cur_node.get_ind()] = true;

        auto [i, k] = cur_node;

        if(i == H) {
            ans = min(ans, cur_time);
            continue;
        }

        for(auto [j, w] : adj[i]) {
            q.push({(arr[j] == 0 ? 0.0 : cur_time + (double)w), {j, k}});
        }

        if(arr[i] == 2 && k <= K) {
            for(auto [j, w] : adj[i]) {
                q.push({cur_time / 2.0 + (double)w, {j, k + 1}});
            }
        }
    }

    if(ans < numeric_limits<double>::max()) {
        return ans;
    }

    return -1;
}

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