제출 #623292

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
623292		BhavayGoyal	Robot (JOI21_ho_t4)	C++14	100 / 100	1162 ms	103416 KiB

Main

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;

const ll INF = 1e18;

struct Edge {
    int to, c;
    ll p;
};

map<int, vector<Edge>> graph[100001];
ll dp[100001];
map<int, ll> dp2[100001], psum[100001];

int main() {
    cin.tie(0)->sync_with_stdio(0);
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= m; i++) {
        int u, v, c;
        ll p;
        cin >> u >> v >> c >> p;
        graph[u][c].push_back({v, c, p});
        graph[v][c].push_back({u, c, p});
        psum[u][c] += p;
        psum[v][c] += p;
    }

    memset(dp, 0x3f, sizeof dp);
    dp[1] = 0;
    priority_queue<tuple<ll, int, int>> pq;
    pq.push({0, 1, 0});
    map<int, int> vis[n+1];
    while (pq.size()) {
        ll cost;
        int node, c;
        tie(cost, node, c) = pq.top();
        pq.pop();
        if (vis[node][c]) continue;
        vis[node][c] = true;

        if (c) {
            if (dp2[node][c] != -cost) continue;
            for (Edge i : graph[node][c]) {
                // We can't flip i in this case
                ll case1 = psum[node][c] - i.p;
                if (case1 - cost < dp[i.to]) {
                    dp[i.to] = case1 - cost;
                    pq.push({-dp[i.to], i.to, 0});
                }
            }
        } else {
            if (dp[node] != -cost) continue;
            for (auto& i : graph[node]) {
                for (Edge j : i.second) {
                    
                    // Case 1: We don't flip j
                    ll case1 = psum[node][j.c] - j.p - cost;

                    // Case 2: We flip j but not another edge of the same colour
                    ll case2 = j.p - cost;

                    // taking min of both cases
                    ll mnCase = min(case1, case2);
                    dp[j.to] = min(mnCase, dp[j.to]);
                    pq.push({-dp[j.to], j.to, 0});
                    
                    // Case 3: We flip j and another edge of the same colour
                    ll case3 = -cost;
                    if (!dp2[j.to].count(j.c)) {
                        dp2[j.to][j.c] = case3;
                        pq.push({-dp2[j.to][j.c], j.to, j.c});
                    }
                    else {
                        dp2[j.to][j.c] = min(case3, dp2[j.to][j.c]);
                        pq.push({-dp2[j.to][j.c], j.to, j.c});
                    }
                }
            }
        }
    }
    cout << (dp[n] > INF ? -1 : dp[n]);
    return 0;
}

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