답안 #982611

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
982611 2024-05-14T13:37:02 Z PanosPask Sailing Race (CEOI12_race) C++14
90 / 100
3000 ms 5052 KB
/*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()); j++) {
                    int n = adj_list[i][j];
                    if (n < N + l) {
                        v = max(v, 1 + dp[MOD(n, N)][MOD(l - 1 + N, N)][0]);
                    }
                    if (n > r_actual) {
                        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

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);
      |             ~~~~~^~~~~~~~~~
# 결과 실행 시간 메모리 Grader output
1 Correct 0 ms 348 KB Output is correct
2 Correct 1 ms 600 KB Output is correct
3 Correct 1 ms 604 KB Output is correct
4 Correct 1 ms 604 KB Output is correct
5 Correct 2 ms 2652 KB Output is correct
6 Correct 4 ms 2652 KB Output is correct
7 Correct 3 ms 2652 KB Output is correct
8 Correct 7 ms 2648 KB Output is correct
9 Correct 6 ms 2652 KB Output is correct
10 Correct 5 ms 2908 KB Output is correct
11 Correct 8 ms 2652 KB Output is correct
12 Correct 121 ms 3340 KB Output is correct
13 Correct 258 ms 3928 KB Output is correct
14 Correct 285 ms 4188 KB Output is correct
15 Correct 1864 ms 4876 KB Output is correct
16 Correct 2575 ms 5052 KB Output is correct
17 Correct 1885 ms 4868 KB Output is correct
18 Correct 545 ms 4696 KB Output is correct
19 Execution timed out 3020 ms 4952 KB Time limit exceeded
20 Execution timed out 3011 ms 4952 KB Time limit exceeded