제출 #1153387

#제출 시각아이디문제언어결과실행 시간메모리
1153387TroySer사이버랜드 (APIO23_cyberland)C++20
44 / 100
110 ms16968 KiB
#include <bits/stdc++.h>

using namespace std;
using ll = long long;

vector<vector<ll> > adjList;
map<pair<ll, ll>, double> weightMat;
vector<bool> isReachable;

void dfs(ll node, ll H, const vector<vector<ll>>& adjList, vector<bool>& isReachable) {

    if (node == H) return;

    for (auto v : adjList[node]) {
        if (!isReachable[v]) {
            isReachable[v] = true;
            dfs(v, H, adjList, isReachable);
        }
    }
}

double dijkstras(ll H, vector<int> &A) {

    ll N = adjList.size();
    
    priority_queue<pair<double, ll>, vector<pair<double, ll> >, greater<pair<double, ll> > > pq;
    pq.push({0.0, H});

    vector<double> distances(N, 1e16);
    vector<bool> visited(N, false);
    distances[H] = 0;

    while (!pq.empty()) {

        auto [currentDistance, currentNode] = pq.top();
        pq.pop();

        if (visited[currentNode]) continue;
        visited[currentNode] = true;

        for (ll v: adjList[currentNode]) {
            double newDist = currentDistance + weightMat[{currentNode, v}];
            if (newDist < distances[v]) {
                distances[v] = newDist;
                pq.push({newDist, v});
            }
        }

    }

    double minimumPossible = distances[0];
    for (ll i = 0; i < A.size(); i++) {
        if ((A[i] == 0) && isReachable[i]) 
            minimumPossible = min(minimumPossible, distances[i]);
    }

    return minimumPossible;

}

double solve(int N, int M, int K, int H, vector<int> x, vector<int> y, vector<int> c, vector<int> arr) {

    adjList.clear();
    weightMat.clear();

    adjList.resize(N);

    for (ll i = 0; i < M; i++) {
        adjList[x[i]].push_back(y[i]);
        adjList[y[i]].push_back(x[i]);
        weightMat[{x[i], y[i]}] = (double)c[i];
        weightMat[{y[i], x[i]}] = (double)c[i];
    }

    isReachable.clear();
    isReachable.resize(N, false);

    dfs(0, H, adjList, isReachable);
    isReachable[0] = true;

    if (!isReachable[H]) {
        return -1;
    }

    double response = dijkstras(H, arr);

    return response;

}
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