Submission #1187674

#TimeUsernameProblemLanguageResultExecution timeMemory
1187674Thanhs사이버랜드 (APIO23_cyberland)C++20
100 / 100
1067 ms64584 KiB
#pragma GCC opmize("Ofast", unroll-loops)
#pragma GCC target("arch=skylake")
#include <bits/stdc++.h>
using namespace std;

#define name "TENBAI"
#define fi first
#define se second
#define endl '\n'
#define setmin(x, y) x = min((x), (y))
#define setmax(x, y) x = max((x), (y))
#define sqr(x) ((x) * (x))
#define all(x) x.begin(), x.end()

mt19937 hdp(chrono::high_resolution_clock::now().time_since_epoch().count());
int rand(int l, int r){return l + ((hdp() % (r - l + 1)) + r - l + 1) % (r - l + 1);}

const double inf = 1 / .0;
const int NM = 1e5 + 5;

double dist[NM][71];

double solve(int N, int M, int K, int H, vector<int> x, vector<int> y, vector<int> c, vector<int> a)
{
    setmin(K, 70);
    fill(dist[0], dist[0] + N * 71, inf);
    vector<vector<pair<int, int>>> g(N);
    for (int i = 0; i < M; i++)
        g[x[i]].push_back({y[i], c[i]}), g[y[i]].push_back({x[i], c[i]});
    vector<int> st(1, 0);
    vector<bool> vis(N, 0);
    priority_queue<tuple<int, double, int>, vector<tuple<int, double, int>>, greater<>> pq;
    while (st.size())
    {
        int u = st.back();
        st.pop_back();
        if (u == H)
            continue;
        vis[u] = 1;
        if (!a[u] || !u)
        {
            for (int i = 0; i <= K; i++)
                dist[u][i] = 0;
            pq.push({0, 0, u});
        }
        for (auto v : g[u])
            if (!vis[v.fi])
            {
                st.push_back(v.fi);
                vis[v.fi] = 1;
            }
    }
    while (pq.size())
    {
        auto u = get<2>(pq.top()), t = get<0>(pq.top());
        double d = get<1>(pq.top());
        pq.pop();
        if (dist[u][t] != d || u == H)
            continue;
        for (pair<int, int> v : g[u])
        {
            if (dist[v.fi][t] > d + v.se)
            {
                dist[v.fi][t] = d + v.se;
                pq.push({t, dist[v.fi][t], v.fi});
            }
            if (a[v.fi] == 2 && t + 1 <= K && dist[v.fi][t + 1] > (d + v.se) / 2.0)
            {
                dist[v.fi][t + 1] = (d + v.se) / 2.0;
                pq.push({t + 1, dist[v.fi][t + 1], v.fi});
            }
        }
    }
    double res = inf;
    for (int i = 0; i <= K; i++)
        setmin(res, dist[H][i]);
    if (res == inf)
        res = -1;
    return res;
}
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