Submission #1184495

#	Time	Username	Problem	Language	Result	Execution time	Memory
1184495		vux2code	Sailing Race (CEOI12_race)	C++20	0 / 100	364 ms	4636 KiB

race

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

static const int MAXN = 500;
int N, K;
vector<int> adj[MAXN];
int dp2[MAXN][MAXN][2]; // classical DP: max edges from st to en, no crossings
int dp1[MAXN][MAXN][2]; // one-crossing DP: max edges from st to en, at most one crossing of initial chord

// Check if x is strictly inside the arc from st to en in direction dir (0=CCW,1=CW)
inline bool inrange(int st, int en, int dir, int x) {
    if (x == st || x == en) return false;
    if (dir == 0) {
        int d_se = (en - st + N) % N;
        int d_sx = (x - st + N) % N;
        return (d_sx > 0 && d_sx < d_se);
    } else {
        int d_se = (st - en + N) % N;
        int d_sx = (st - x + N) % N;
        return (d_sx > 0 && d_sx < d_se);
    }
}

// Classical DP (k=0): longest path (in edges) from st to en without crossing the chord en->st
int dfs2(int st, int en, int dir) {
    int &res = dp2[st][en][dir];
    if (res != -1) return res;
    int best = 0;
    for (int nxt : adj[st]) {
        if (nxt == en || inrange(st, en, dir, nxt)) {
            best = max(best, 1 + dfs2(nxt, en, dir));
        }
    }
    return res = best;
}

// One-crossing DP (k=1): longest path (in edges) from st to en allowing at most one crossing of chord en->st
int dfs1(int st, int en, int dir) {
    int &res = dp1[st][en][dir];
    if (res != -1) return res;
    if (st == en) return res = 0;
    // Option 1: never cross
    int best = dfs2(st, en, dir);
    // Option 2: make a move; if it crosses, switch to classical DP on opposite arc
    for (int nxt : adj[st]) {
        if (nxt == en || inrange(st, en, dir, nxt)) {
            best = max(best, 1 + dfs1(nxt, en, dir));
        } else {
            best = max(best, 1 + dfs2(nxt, en, 1 - dir));
        }
    }
    return res = best;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    cin >> N >> K;
    for (int i = 0; i < N; i++) {
        adj[i].clear();
        while (true) {
            int x; cin >> x;
            if (x == 0) break;
            adj[i].push_back(x - 1);
        }
    }

    memset(dp2, -1, sizeof(dp2));
    memset(dp1, -1, sizeof(dp1));

    int answer = 0;
    int startHarbor = 1;
    // Try each possible first stage s->t
    for (int s = 0; s < N; s++) {
        for (int t : adj[s]) {
            for (int dir = 0; dir < 2; dir++) {
                int val;
                if (K == 0) {
                    // classical
                    val = 1 + dfs2(t, s, 1 - dir);
                } else {
                    // allow one crossing
                    val = 1 + dfs1(t, s, 1 - dir);
                }
                if (val > answer) {
                    answer = val;
                    startHarbor = s + 1;
                }
            }
        }
    }

    cout << answer << "\n" << startHarbor << "\n";
    return 0;
}

• Subtask 1: 0 / 100