답안 #982614

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
982614 2024-05-14T13:40:13 Z PanosPask Sailing Race (CEOI12_race) C++14
90 / 100
3000 ms 5060 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() - 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

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 344 KB Output is correct
2 Correct 1 ms 612 KB Output is correct
3 Correct 1 ms 600 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 6 ms 2648 KB Output is correct
9 Correct 5 ms 2652 KB Output is correct
10 Correct 5 ms 2904 KB Output is correct
11 Correct 8 ms 2896 KB Output is correct
12 Correct 115 ms 3164 KB Output is correct
13 Correct 250 ms 4044 KB Output is correct
14 Correct 280 ms 4252 KB Output is correct
15 Correct 1700 ms 4868 KB Output is correct
16 Correct 2333 ms 5060 KB Output is correct
17 Correct 1716 ms 4872 KB Output is correct
18 Correct 576 ms 4700 KB Output is correct
19 Execution timed out 3011 ms 4952 KB Time limit exceeded
20 Execution timed out 3071 ms 4956 KB Time limit exceeded