제출 #1201411

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1201411		zh_h	사이버랜드 (APIO23_cyberland)	C++17	44 / 100	980 ms	2162688 KiB

cyberland

#include<bits/stdc++.h>
#define pb push_back
using namespace std;

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, double>>> edge(N);
    for (int i = 0; i < M; i ++) {
        edge[x[i]].pb({y[i], c[i]});
        edge[y[i]].pb({x[i], c[i]});
    }

    vector<bool> visited(N, false);
    queue<int> q;
    q.push(0);
    while (!q.empty()) {
        int v = q.front(); q.pop();
        
        if (v == H) continue;
        for (auto [u, w] : edge[v]) {
            if (visited[u]) continue;
            visited[u] = true;
            q.push(u);
        }
    }

    priority_queue<pair<double, pair<int, int>>> pq;
    vector<vector<double>> dist(N, vector<double>(K+1, 1e19));
    dist[0][0] = 0;
    pq.push({0, {0, 0}}); // {-w, {v, k}}
    
    // cout << "zeros: ";
    for (int i = 0; i < N; i ++) {
        if (arr[i] == 0 && visited[i]) {
            // cout << i << " ";
            pq.push({0, {i, 0}});
            dist[i][0] = 0;
        }
    }
    
    // cout << endl;
    
    visited.clear();
    visited.resize(N, false);
    
    // vector<vector<double>> dist(N, vector<double>(31, 1e19));
    // dp[i][t] = min cost to reach to node i with t times of power 2 ability used

    while (!pq.empty()) {
        int v = pq.top().second.first, k = pq.top().second.second; pq.pop();
        // cout << v << endl;
        visited[v] = true;

        // priority_queue<pair<double, int>> pq_temp = pq;
        // while (!pq_temp.empty()) {
        //     cout << pq_temp.top().first << " " << pq_temp.top().second << endl;
        //     pq_temp.pop();
        // }
        // cout << endl;
        // for (auto i : visited) {cout << i << " ";} cout << endl;

        for (auto [u, w] : edge[v]) {
            if (visited[u]) continue;

            if (dist[v][k] + w < dist[u][k]) {
                dist[u][k] = dist[v][k] + w;
                // if (arr[u] == 0) {dist[u] = 0;}
                pq.push({ (dist[u][k]*(-1)) , {u, k}});

            }

            if (arr[u] == 2 && k+1 <= K) {
                if (dist[v][k+1] + w < dist[u][k+1]) {
                    dist[u][k+1] = dist[v][k+1] + w;
                    pq.push({ (dist[u][k+1]*(-1)), {u, k+1}});
                }
            }
        }

    }

    // if (dist[0] == 1e19) {return (double)-1;}
    // return (double)dist[0];
    double mn = 1e19;
    for (int i = 0; i <= K; i ++) {
        mn = min(mn, dist[H][i]);
    }
    if (mn == 1e19) {return (double)-1;}
    return (double)mn;
}

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