제출 #1191278

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1191278		just	사이버랜드 (APIO23_cyberland)	C++20	5 / 100	39 ms	13128 KiB

cyberland

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

#define vec vector
#define all(x) (x).begin(), (x).end()

double solve(int N, int M, int K, int H, vec<int> X, vec<int> Y, vec<int> C, vec<int> S) {
    using pii = pair<int, int>;

    map<pii, int> edges;
    vec<vec<pii>> adj(N);
    for (int i = 0; i < M; i++) {
        int u = X[i], v = Y[i], cost = C[i];
        adj[u].push_back({v, cost});
        adj[v].push_back({u, cost});

        edges[{u, v}] = cost;
        edges[{v, u}] = cost;
    }

    vec<int> dist(N, INT_MAX);
    dist[0] = 0;
    priority_queue<pii, vec<pii>, greater<pii>> pq;
    pq.push({0, 0});
    while (!pq.empty()) {
        auto [d, u] = pq.top();
        pq.pop();
        if (d > dist[u]) continue;
        for (auto [v, cost] : adj[u]) {
            if (dist[v] > d + cost) {
                dist[v] = d + cost;
                pq.push({dist[v], v});
            }
        }
    }

    if (dist[H] == INT_MAX) return -1;

    if (N == 2) {
        return dist[H];
    }

    // Subtask O1
    if (N == 3) {
        int other = (H == 1 ? 2 : 1);
        double result = INT_MAX;
        if (edges.count({0, H})) {
            result = edges[{0, H}];
        }

        if (edges.count({0, other}) && edges.count({other, H})) {
            if (S[other] == 0)
                result = min(result, (double)edges[{other, H}]);
            if (S[other] == 1)
                result = min(result, (double)dist[H]);
            if (S[other] == 2)
                result = min(result, (double)edges[{0, other}] / 2.0 + edges[{other, H}]);
        }

        return result;
    }

    vec<int> backdist(N, INT_MAX);
    backdist[H] = 0;
    priority_queue<pii, vec<pii>, greater<pii>> backpq;

    backpq.push({0, H});
    while (!backpq.empty()) {
        auto [d, u] = backpq.top();
        backpq.pop();
        if (d > backdist[u]) continue;
        for (auto [v, cost] : adj[u]) {
            if (backdist[v] > d + cost) {
                backdist[v] = d + cost;
                backpq.push({backdist[v], v});
            }
        }
    }
    if (backdist[0] == INT_MAX) return -1;

    return backdist[H];
}

• Subtask 1: 5 / 5
• Subtask 2: 0 / 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