제출 #1158519

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1158519		countless	사이버랜드 (APIO23_cyberland)	C++20	63 / 100	944 ms	2162688 KiB

cyberland

#include "cyberland.h"

#include <vector>
#include <queue>
#include <iostream>

using namespace std;

typedef long long ll;
typedef long double ld;

const ld INF = 9e18;

// hmm okay this looks like it works
struct T {
    ld cdist;
    ll k;
    ll node;

    bool operator>(const T &other) const {
        if (cdist != other.cdist) return cdist > other.cdist;
        if (k != other.k) return k > other.k;
        return node > other.node;
    }
};

// this should work for k <= 30
ld divByPow2(ll x, ll k) {
    if (k > 30) return (ld)0;

    return (ld)(x / (1LL << k));
}

// pretty sure this works too
void checkReal(int start, int H, vector<vector<pair<ll, ll>>> &adj, vector<bool> &real) {
    real[start] = true;

    if (start == H) return;

    for (const pair<ll, ll> &i : adj[start]) {
        if (!real[i.first]) {
            checkReal(i.first, H, adj, real);
        }
    }
}

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<ll, ll>>> adj(N+5);

    for (int i = 0; i < M; i++) {
        adj[x[i]].push_back({y[i], c[i]});
        adj[y[i]].push_back({x[i], c[i]});
    }

    vector<bool> real(N+5, false);
    checkReal(0, H, adj, real);

    if (!real[H]) return -1;

    vector<vector<ld>> dist(N+5, vector<ld>(K + 1 + 5, INF));
    dist[H][0] = 0;
    priority_queue<T, vector<T>, greater<T>> pq;
    pq.push({0, 0, H});

    while (!pq.empty()) {
        const auto [cdist, k, node] = pq.top();
        pq.pop();

        // if (cdist > dist[node][k]) continue;

        if ((arr[node] == 0 || node == 0) && real[node]) {
            return cdist;
        }

        if (cdist != dist[node][k]) continue;
        for (const pair<ll, ll> &i : adj[node]) {
            if (i.first == H) continue;
            if (!real[i.first]) continue;

            if ((cdist + divByPow2(i.second, k)) < dist[i.first][k]) {
                dist[i.first][k] = cdist + divByPow2(i.second, k);
                pq.push({dist[i.first][k], k, i.first});
            }

            if (arr[node] == 2 && k < K && (cdist + divByPow2(i.second, k + 1)) < dist[i.first][k+1]) {
                dist[i.first][k+1] = cdist + divByPow2(i.second, k + 1);
                pq.push({dist[i.first][k+1], k + 1, i.first});
            }
        }
    }

    return -1;
}

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