제출 #1228103

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1228103		Dzadzo	사이버랜드 (APIO23_cyberland)	C++20	15 / 100	860 ms	41372 KiB

cyberland

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

// Use priority_queue instead of set for Dijkstra's algorithm
// State: (node, number of half-cost moves used)

double solve(int n, int m, int K, int h,
             vector<int> x,  vector<int> y,
              vector<int> c,  vector<int> arr) {
    vector<vector<pair<int,int>>> adj(n);
    for (int i = 0; i < m; i++) {
        adj[x[i]].emplace_back(y[i], c[i]);
        adj[y[i]].emplace_back(x[i], c[i]);
    }

    // Check reachability of h from 0
    vector<bool> visited(n, false);
    function<void(int)> dfs = [&](int v) {
        if (visited[v]) return;
        visited[v] = true;
        for (auto& [to, w] : adj[v]) {
            dfs(to);
        }
    };
    dfs(0);
    if (!visited[h]) return -1;

    const double INF = 1e15;
    int maxK = min(K, 30);
    vector<vector<double>> dist(n, vector<double>(maxK + 1, INF));

    // Initialize distances: start at 0 with 0 half-cost moves
    priority_queue<
        pair<double, pair<int,int>>,
        vector<pair<double, pair<int,int>>>,
        greater<>> pq;

    dist[0][0] = 0;
    pq.push({0.0, {0, 0}});

    // Also, if arr[i]==0 and reachable, they can be considered as extra start points with zero cost
    for (int i = 1; i < n; i++) {
        if (!arr[i] && visited[i]) {
            dist[i][0] = 0;
            pq.push({0.0, {i, 0}});
        }
    }

    while (!pq.empty()) {
        auto [d, state] = pq.top();
        pq.pop();
        int v = state.first;
        int used = state.second;
        if (d > dist[v][used]) continue;
        if (v == h) continue;

        for (auto& [to, w] : adj[v]) {
            // Regular edge
            if (dist[to][used] > d + w) {
                dist[to][used] = d + w;
                pq.push({dist[to][used], {to, used}});
            }
            // Half-cost edge if allowed and special node
            if (arr[to] == 2 && used < maxK) {
                double nd = (d + w) / 2.0;
                if (dist[to][used + 1] > nd) {
                    dist[to][used + 1] = nd;
                    pq.push({nd, {to, used + 1}});
                }
            }
        }
    }

    double ans = INF;
    for (int i = 0; i <= maxK; i++) {
        ans = min(ans, dist[h][i]);
    }
    return (ans >= INF ? -1 : ans);
}

• 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