제출 #400963

#제출 시각아이디문제언어결과실행 시간메모리
400963534351From Hacks to Snitches (BOI21_watchmen)C++17
15 / 100
6097 ms88816 KiB
#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 int MAXN = 250013;
const int MAXC = 2813;
const int INF = 1e9 + 7;
const int DELTA = 125;

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, K, B;
vi edge[MAXN];
int ban[MAXN], md[MAXN], id[MAXN], dist[MAXN], cc[MAXN];
set<pii> q;

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;
        cin >> u >> v; u--; v--;
        edge[u].PB(v);
        edge[v].PB(u);
    }
    FOR(i, 0, N)
    {
        edge[i].PB(i);
        cc[i] = -1;
    }
    cin >> K;
    FOR(i, 0, K)
    {
        int c; cin >> c;
        vi s(c);
        FOR(j, 0, c)
        {
            cin >> s[j]; s[j]--;
            id[s[j]] = B; B++;
            ban[s[j]] = j;
            cc[s[j]] = i;
        }
        md[i] = c;
    }
    fill(dist, dist + N, INF);
    dist[0] = 0;
    q.insert({0, 0});
    while(!q.empty())
    {
        int d = q.begin() -> fi, u = q.begin() -> se; q.erase(q.begin());
        if (u == N - 1) break;
        // cerr << d << ' ' << u << endl;
        // bool ok = false;
        // for (int v : edge[u])
        // {
        //     if (!seen[v][0]) ok = true;
        //     //make sure there's one univisited neighbor
        // }
        // if (!ok) continue;
        for (int v : edge[u])
        {
            if (cc[v] != -1)
            {
                FOR(j, 1, (cc[u] == -1 ? DELTA : 2))
                {
                    // cerr << u << ' ' << v << ' ' << (d + j) << endl;
                    int m = md[cc[v]];
                    if ((d + j) % m == ban[v]) continue;
                    if (cc[u] == cc[v] && (d + j) % m == ban[u] && (d + j - 1) % m == ban[v]) continue;
                    if (dist[v] > d + j)
                    {
                        dist[v] = d + j;
                    }
                    if (d + j - dist[v] >= DELTA) continue;
                    q.insert({d + j, v});
                }
            }
            else
            {
                if (dist[v] > d + 1)
                {
                    dist[v] = d + 1;
                    q.insert({d + 1, v});
                }
            }
        }
    }
    if (dist[N - 1] >= INF)
    {
        cout << "impossible\n";
        return 0;
    }
    cout << dist[N - 1] << '\n';
    return 0;
}
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