제출 #222603

#제출 시각아이디문제언어결과실행 시간메모리
222603dolphingarlicSailing Race (CEOI12_race)C++14
100 / 100
1552 ms6780 KiB
#include <bits/stdc++.h>
#define FOR(i, x, y) for (int i = x; i < y; i++)
typedef long long ll;
using namespace std;

int dp[500][500][2], dist[500][500][2], best_range[500][500][2];
bool adj[500][500];

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int n, t;
    cin >> n >> t;
    FOR(i, 0, n) FOR(j, 0, n) FOR(k, 0, 2) dist[i][j][k] = -n;
    FOR(i, 0, n) {
        int v;
        cin >> v;
        while (v) {
            adj[i][v - 1] = true;
            dist[i][v - 1][0] = dist[v - 1][i][1] = 1;
            cin >> v;
        }
    }

    FOR(k, 1, n + 1) FOR(i, 0, n) {
        int j = (i + k) % n;
        FOR(v, 0, n) {
            if ((v > i && v < i + k) || (v > i - n && v < i + k - n)) {
                if (adj[i][v]) dp[i][j][0] = max(dp[i][j][0], max(dp[v][j][0], dp[i][v][1]) + 1);
                if (adj[j][v]) dp[i][j][1] = max(dp[i][j][1], max(dp[v][j][0], dp[i][v][1]) + 1);
                dist[i][j][0] = max(dist[i][j][0], dist[i][v][0] + dist[v][j][0]);
                dist[i][j][1] = max(dist[i][j][1], dist[i][v][1] + dist[v][j][1]);
            }
        }
        best_range[i][j][0] = max(best_range[(i + 1) % n][j][0], dp[i][j][0]);
        best_range[i][j][1] = max(best_range[i][(j - 1 + n) % n][1], dp[i][j][1]);
    }

    pair<int, int> ans = {0, -1};
    FOR(i, 0, n) ans = max(ans, {max(dp[i][i][0], dp[i][i][1]), i});
    if (t) {
        FOR(k, 1, n + 1) FOR(i, 0, n) {
            int j = (i + k) % n;
            FOR(v, i + k + 1, i + n) if (adj[v % n][i]) {
                FOR(x, v + 1, i + n) if (adj[j][x % n]) {
                    ans = max(ans, {dist[i][j][0] + max(best_range[v % n][x % n][1], best_range[x % n][i][0]) + 2, v % n});
                }
                break;
            }
            for (int v = i + n - 1; v > i + k; v--) if (adj[v % n][j]) {
                for (int x = v - 1; x > i + k; x--) if (adj[i][x % n]) {
                    ans = max(ans, {dist[i][j][1] + max(best_range[x % n][v % n][0], best_range[j][x % n][1]) + 2, v % n});
                }
                break;
            }
        }
    }
    cout << ans.first << '\n' << ans.second + 1 << '\n';
    return 0;
}
#Verdict Execution timeMemoryGrader output
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