제출 #485260

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
485260		arujbansal	Robot (JOI21_ho_t4)	C++17	100 / 100	968 ms	83776 KiB

Main

#include <bits/stdc++.h>

using namespace std;

using ll = long long;

const int MXN = 1e5 + 1, MXM = 2e5 + 1;
const ll INF = 1e18 + 5;

struct state {
    int node, colour;
    ll cur_dist;

    state(int _node = 0, int _colour = 0, ll _cur_dist = 0) : node(_node), colour(_colour), cur_dist(_cur_dist) {}

    bool operator<(const state &other) const {
        return cur_dist > other.cur_dist;
    }
};

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int N, M;
    cin >> N >> M;

    map<int, ll> sum[N];
    vector<array<int, 3>> g[N];
    map<int, vector<pair<int, int>>> edges[N];

    for (int i = 0; i < M; i++) {
        int a, b, c, p;
        cin >> a >> b >> c >> p;
        a--, b--;

        // {v, colour of edge, cost of deleting edge}
        g[a].emplace_back(array<int, 3>{b, c, p});
        g[b].emplace_back(array<int, 3>{a, c, p});

        // Sum of costs of all edges incident to a node with a particular colour
        sum[a][c] += p;
        sum[b][c] += p;

        // All edges incident to a node with a particular colour
        edges[a][c].emplace_back(b, p);
        edges[b][c].emplace_back(a, p);
    }

    vector<ll> dist(N, INF);
    dist[0] = 0;

    map<int, ll> dist_col[N];

    priority_queue<state> pq;
    pq.emplace(0, -1, 0); // -1 represents free will, no colour restriction

    while (!pq.empty()) {
        auto [u, colour, cur_dist] = pq.top();
        pq.pop();

        if (colour == -1) {
            if (dist[u] != cur_dist) continue;

            for (const auto &[v, edge_col, edge_cost] : g[u]) {
                // 2.1 and 2.2
                ll cost1 = min(cur_dist + sum[u][edge_col] - edge_cost, cur_dist + edge_cost);
                if (cost1 < dist[v]) {
                    dist[v] = cost1;
                    pq.emplace(v, -1, dist[v]);
                }

                // 2.3
                if (!dist_col[v].count(edge_col) || cur_dist < dist_col[v][edge_col]) {
                    dist_col[v][edge_col] = cur_dist;
                    pq.emplace(v, edge_col, cur_dist);
                }
            }
        } else {
            if (cur_dist != dist_col[u][colour]) continue;

            for (const auto &[v, edge_cost] : edges[u][colour]) {
                ll cost = cur_dist + sum[u][colour] - edge_cost;
                if (cost < dist[v]) {
                    dist[v] = cost;
                    pq.emplace(v, -1, cost);
                }
            }
        }
    }

    cout << (dist[N - 1] < INF ? dist[N - 1] : -1);
}

/*
    Observations:
    1. Always possible to change the colour of an edge so that it is different from all other edges connecting u if the answer is not -1 (?)
    2. Three cases if current edge E(u, v) is of colour C:
        2.1: We use E by changing colours of all other edges with C to go to v. At v, it's okay to use an edge of colour C since it gives us a worse cost than 2.2.
        2.2: We change E's colour and go to v. Don't change the colour of any other edge incident to u with colour C. We can use an edge of any colour from v.
             This gives us a worse cost if we use an edge of colour C form v because we double count the cost for this edge. Case 2.3 handles this.
        2.3: We change E's colour and use an edge of colour C from v.
*/

• Subtask 1: 34 / 34
• Subtask 2: 24 / 24
• Subtask 3: 42 / 42