Submission #831464

#	Time	Username	Problem	Language	Result	Execution time	Memory
831464		maomao90	Giraffes (JOI22_giraffes)	C++17	59 / 100	7059 ms	624 KiB

giraffes

// Hallelujah, praise the one who set me free
// Hallelujah, death has lost its grip on me
// You have broken every chain, There's salvation in your name
// Jesus Christ, my living hope
#include <bits/stdc++.h> 
using namespace std;

#define REP(i, s, e) for (int i = (s); i < (e); i++)
#define RREP(i, s, e) for (int i = (s); i >= (e); i--)
template <class T>
inline bool mnto(T& a, T b) {return a > b ? a = b, 1 : 0;}
template <class T>
inline bool mxto(T& a, T b) {return a < b ? a = b, 1: 0;}

typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;
#define FI first
#define SE second
typedef pair<int, int> ii;
typedef pair<ll, ll> pll;
typedef tuple<int, int, int> iii;
#define ALL(_a) _a.begin(), _a.end()
#define SZ(_a) (int) _a.size()
#define pb push_back
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<ii> vii;
typedef vector<iii> viii;

#ifndef DEBUG
#define cerr if (0) cerr
#endif

const int INF = 1000000005;
const ll LINF = 1000000000000000005ll;
const int MAXN = 8005;

int n;
int p[MAXN];
//int dp[MAXN][MAXN][MAXN];
viii pdp, ndp;

int main() {
#ifndef DEBUG
    ios::sync_with_stdio(0), cin.tie(0);
#endif
    cin >> n;
    REP (i, 1, n + 1) {
        cin >> p[i];
        p[i]--;
    }
    pdp.pb({1, n, 0});
    REP (x, 0, n + 1) {
        ndp.clear();
        REP (l, 1, n + 1) { // p[l] is min
            int r = -INF;
            for (auto [pl, pr, pmn] : pdp) {
                if (l < pl || l > pr) {
                    continue;
                }
                int pmx = pmn + pr - pl;
                if (p[l] < pmn || p[l] > pmx) {
                    continue;
                }
                int tr = pr - max(0, p[l] - pmn - (l - pl));
                mxto(r, tr);
            }
            if (r != -INF) {
                ndp.pb({l + 1, r, p[l] + 1});
            }
        }
        REP (l, 1, n + 1) { // p[l] is max
            int r = -INF;
            for (auto [pl, pr, pmn] : pdp) {
                if (l < pl || l > pr) {
                    continue;
                }
                int pmx = pmn + pr - pl;
                if (p[l] < pmn || p[l] > pmx) {
                    continue;
                }
                int tr = pr - max(0, pmx - p[l] - (l - pl));
                mxto(r, tr);
            }
            if (r != -INF) {
                ndp.pb({l + 1, r, p[l] - (r - l)});
            }
        }
        REP (r, 1, n + 1) { // p[r] is min
            int l = INF;
            for (auto [pl, pr, pmn] : pdp) {
                if (r < pl || r > pr) {
                    continue;
                }
                int pmx = pmn + pr - pl;
                if (p[r] < pmn || p[r] > pmx) {
                    continue;
                }
                int tl = pl + max(0, p[r] - pmn - (pr - r));
                mnto(l, tl);
            }
            if (l != INF) {
                ndp.pb({l, r - 1, p[r] + 1});
            }
        }
        REP (r, 1, n + 1) { // p[r] is max
            int l = INF;
            for (auto [pl, pr, pmn] : pdp) {
                if (r < pl || r > pr) {
                    continue;
                }
                int pmx = pmn + pr - pl;
                if (p[r] < pmn || p[r] > pmx) {
                    continue;
                }
                int tl = pl + max(0, pmx - p[r] - (pr - r));
                mnto(l, tl);
            }
            if (l != INF) {
                ndp.pb({l, r - 1, p[r] - (r - l)});
            }
        }
        if (ndp.empty()) {
            cout << n - x << '\n';
            return 0;
        }
        swap(ndp, pdp);
    }
    /*
    RREP (l, n, 1) {
        REP (r, l, n + 1) {
            REP (mn, 0, n) {
                int mx = mn + (r - l);
                dp[l][r][mn] = max({
                        dp[l + 1][r][mn + 1] + (p[l] == mn),
                        dp[l + 1][r][mn] + (p[l] == mx),
                        dp[l][r - 1][mn + 1] + (p[r] == mn),
                        dp[l][r - 1][mn] + (p[r] == mx)});
            }
        }
    }
    cout << n - dp[1][n][0] << '\n';
    */
    return 0;
}

• Subtask 1: 10 / 10
• Subtask 2: 22 / 22
• Subtask 3: 27 / 27
• Subtask 4: 0 / 41