Submission #982614

#	Time	Username	Problem	Language	Result	Execution time	Memory
982614		PanosPask	Sailing Race (CEOI12_race)	C++14	90 / 100	3071 ms	5060 KiB

race

/*This problem needs arrays instad of vectors due to
extremely tight ML and TL*/

#include <bits/stdc++.h>
#define pb push_back
#define MOD(var, M) (((var) >= (M)) ? ((var) - M) : (var))

using namespace std;

const int MAXN = 500;
const int INF = 1e9;

int N, K;
bool stage[MAXN][MAXN];

vector<vector<int>> adj_list;

// dp[l][r][k]:  Maximum number of stages if enclosed by l, r
// and being in the stages such that l < s < r'
// Where r' = l > r ? r = r + N : r

// k == 0: Starting in l
// k == 1: Starting in r
int dp[MAXN][MAXN][2];

// Before the intersection with the first stage
// You can only choose one way to follow
// Either from l to r or from r to l
int bef[MAXN][MAXN][2];

int main(void)
{
    scanf("%d %d", &N, &K);

    adj_list.resize(N);

    for (int i = 0; i < N; i++) {
        int v;
        scanf("%d", &v);

        while (v != 0) {
            v--;
            stage[i][v] = true;
            adj_list[i].pb(v);
            adj_list[i].pb(v + N);
            scanf("%d", &v);
        }

        sort(adj_list[i].begin(), adj_list[i].end());
    }

    for (int len = 1; len <= N; len++) {
        for (int l = 0; l < N; l++) {
            int r_actual = l + len - 1;
            int r = MOD(r_actual, N);

            bef[l][r][0] = bef[l][r][1] = -INF;

            for (int i_actual = l + 1; i_actual < r_actual; i_actual++) {
                int i = MOD(i_actual, N);

                dp[l][r][0] = max(dp[l][r][0], dp[l][i][0]);
                dp[l][r][1] = max(dp[l][r][1], dp[i][r][1]);

                if (bef[l][i][0] != -INF && bef[i][r][0] != -INF) {
                    bef[l][r][0] = max(bef[l][r][0], bef[l][i][0] + bef[i][r][0]);
                }
                if (bef[l][i][1] != -INF && bef[i][r][1] != -INF) {
                    bef[l][r][1] = max(bef[l][r][1], bef[l][i][1] + bef[i][r][1]);
                }

                if (stage[l][i]) {
                    dp[l][r][0] = max(dp[l][r][0], 1 + dp[i][r][0]);
                }
                if (stage[r][i]) {
                    dp[l][r][1] = max(dp[l][r][1], 1 + dp[l][i][1]);
                }
            }

            if (stage[l][r]) {
                int res = 1 + dp[MOD(l + 1, N)][r][1];
                if (res >= dp[l][r][0]) {
                    dp[l][r][0] = res;
                }

                bef[l][r][0] = max(bef[l][r][0], 1);
            }
            if (stage[r][l]) {
                int res = 1 + dp[l][MOD(r_actual - 1, N)][0];
                if (res >= dp[l][r][1]) {
                    dp[l][r][1] = res;
                }

                bef[l][r][1] = max(bef[l][r][1], 1);
            }
        }
    }

    int ans = 0;
    int starting = 0;
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
            if (dp[i][j][0] > ans) {
                ans = dp[i][j][0];
                starting = i;
            }
            if (dp[i][j][1] > ans) {
                ans = dp[i][j][1];
                starting = j;
            }
        }
    }
    if (K == 0) {
        printf("%d\n%d\n", ans, starting + 1);
        return 0;
    }

    // Unite bef(before intersection) with dp(after intersection)
    for (int len = 0; len <= N; len++) {
        for (int l = 0; l < N; l++) {
            int r_actual = l + len - 1;
            int r = MOD(r_actual, N);

            bool g1 = stage[l][r];
            bool g2 = stage[r][l];

            for (int i_actual = l + 1; i_actual < r_actual; i_actual++) {
                int i = MOD(i_actual, N);

                int res = -INF;
                int s = -1;
                if (g2) {
                    res = bef[l][i][0];
                    s = r;
                }
                if (g1 && res < bef[i][r][1]) {
                    res = bef[i][r][1];
                    s = l;
                }
                res++;

                if (res < 0) {
                    continue;
                }

                int j1 = upper_bound(adj_list[i].begin(), adj_list[i].end(), r_actual) - adj_list[i].begin();
                int j2 = lower_bound(adj_list[i].begin(), adj_list[i].end(), N + l) - adj_list[i].begin() - 1;

                int v = 0;
                for (int j = max(j1, 0); j <= min(j2, (int)adj_list[i].size() - 1); j++) {
                    int n = adj_list[i][j];
                    v = max(v, 1 + dp[MOD(n, N)][MOD(l - 1 + N, N)][0]);
                    v = max(v, 1 + dp[MOD(r + 1, N)][MOD(n, N)][1]);
                }

                if (ans < v + res) {
                    ans = v + res;
                    starting = s;
                }
            }
        }
    }

    printf("%d\n%d\n", ans, starting + 1);

    return 0;
}

Compilation message (stderr)

race.cpp: In function 'int main()':
race.cpp:33:10: warning: ignoring return value of 'int scanf(const char*, ...)' declared with attribute 'warn_unused_result' [-Wunused-result]
   33 |     scanf("%d %d", &N, &K);
      |     ~~~~~^~~~~~~~~~~~~~~~~
race.cpp:39:14: warning: ignoring return value of 'int scanf(const char*, ...)' declared with attribute 'warn_unused_result' [-Wunused-result]
   39 |         scanf("%d", &v);
      |         ~~~~~^~~~~~~~~~
race.cpp:46:18: warning: ignoring return value of 'int scanf(const char*, ...)' declared with attribute 'warn_unused_result' [-Wunused-result]
   46 |             scanf("%d", &v);
      |             ~~~~~^~~~~~~~~~

• Subtask 1: 90 / 100