Submission #409494

#	Time	Username	Problem	Language	Result	Execution time	Memory
409494		534351	Robot (JOI21_ho_t4)	C++17	100 / 100	1672 ms	114684 KiB

Main

#include <bits/stdc++.h>

using namespace std;

template<class T, class U>
void ckmin(T &a, U b)
{
    if (a > b) a = b;
}

template<class T, class U>
void ckmax(T &a, U b)
{
    if (a < b) a = b;
}

#define MP make_pair
#define PB push_back
#define LB lower_bound
#define UB upper_bound
#define fi first
#define se second
#define SZ(x) ((int) (x).size())
#define ALL(x) (x).begin(), (x).end()
#define FOR(i, a, b) for (auto i = (a); i < (b); i++)
#define FORD(i, a, b) for (auto i = (a) - 1; i >= (b); i--)

const long long LLINF = 3e18;
const int MAXN = 200013;

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpi;
typedef vector<pll> vpl;

int N, M;
map<int, vpi> edge[MAXN];
int color[MAXN];
int cost[MAXN];
map<int, ll> sum[MAXN];
//you can only save it if its color is unique or its color is twice and you just used one of them & paid.
ll dist[MAXN];
map<int, ll> special[MAXN];
priority_queue<array<ll, 3>, vector<array<ll, 3> >, greater<array<ll, 3> > > pq;
ll ans;
//distance, vertex, edge

int32_t main()
{
    cout << fixed << setprecision(12);
    cerr << fixed << setprecision(4);
    ios_base::sync_with_stdio(false); cin.tie(0);
    cin >> N >> M;
    FOR(i, 0, M)
    {
        int u, v, c, d;
        cin >> u >> v >> c >> d; u--; v--;
        edge[u][c].PB({v, i});
        edge[v][c].PB({u, i});
        sum[u][c] += d;
        sum[v][c] += d;
        color[i] = c;
        cost[i] = d;
        // edge[u].PB({v, c, d});
        // edge[v].PB({u, c, d});
    }
    FOR(u, 0, N)
    {
        for (auto &p : edge[u])
        {
            sort(ALL(p.se), [&](pii a, pii b)
            {
                return cost[a.se] > cost[b.se];
            });
            // for (auto e : p.se)
            // {
            //     cerr << "EDGE " << u << ' ' << e.fi << ' ' << " color " << p.fi << " dis " << e.se << endl;
            // }
        }
    }
    fill(dist, dist + N, LLINF);
    dist[0] = 0;
    pq.push({0, 0, -1});
    while(!pq.empty())
    {
        ll d = pq.top()[0]; int u = pq.top()[1], c = pq.top()[2];
        pq.pop();
        if (c != -1)
        {
            if (d != special[u][c]) continue;
            int cl = color[c];
            int it = 0;
            for (auto e : edge[u][cl]) //you can take an edge for free if you just came here w edge of that color && deg is 2.
            {
                // #warning cut time complexity by limiting it to two iterations?
                int v = e.fi, eid = e.se; ll dis = cost[eid];
                ll nd = d + min(dis, sum[u][cl] - dis - cost[c]);
                if (dist[v] > nd)
                {
                    dist[v] = nd;
                    pq.push({nd, v, -1});
                }
                it++;
                if (it > 3) break;
                // if (special[v].find(eid) == special[v].end() || special[v])
            }
        }
        else
        {
            if (d != dist[u]) continue;
            for (auto p : edge[u])
            {
                int cl = p.fi;
                for (auto e : p.se)
                {
                    int v = e.fi, eid = e.se; ll dis = cost[eid];
                    //no guarantees -> you just need to make this the unique color from u.
                    ll nd = d + min(dis, sum[u][cl] - dis);
                    // ll nd = d + dis;
                    if (dist[v] > nd)
                    {
                        dist[v] = nd;
                        pq.push({nd, v, -1});
                    }
                    //but also set the special distance. here you have to actually pay to set this.
                    nd = d + dis;
                    if ((special[v].find(eid) == special[v].end() || special[v][eid] > nd))
                    {
                        special[v][eid] = nd;
                        pq.push({nd, v, eid});
                    }
                }
            }
        }
    }
    ans = dist[N - 1];
    if (ans >= LLINF) ans = -1;
    cout << ans << '\n';
    return 0;
}

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