Submission #831344

#	Time	Username	Problem	Language	Result	Execution time	Memory
831344		maomao90	Giraffes (JOI22_giraffes)	C++17	32 / 100	40 ms	2980 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 = 13;

int n;
int p[MAXN + 5];
int dp[MAXN + 5][MAXN + 5][1 << MAXN];

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]--;
    }
    dp[n + 1][n][0] = 0;
    RREP (l, n, 1) {
        dp[l][l - 1][0] = 0;
        REP (r, l, n + 1) {
            REP (msk, 0, 1 << n) {
                int pop = 0, mn = INF, mx = -INF;
                REP (i, 0, n) {
                    if (msk >> i & 1) {
                        mnto(mn, i);
                        mxto(mx, i);
                        pop++;
                    }
                }
                if (pop != r - l + 1) {
                    continue;
                }
                dp[l][r][msk] = min({
                        dp[l + 1][r][msk ^ (1 << mn)] + (p[l] != mn),
                        dp[l + 1][r][msk ^ (1 << mx)] + (p[l] != mx),
                        dp[l][r - 1][msk ^ (1 << mn)] + (p[r] != mn),
                        dp[l][r - 1][msk ^ (1 << mx)] + (p[r] != mx)});
            }
        }
    }
    cout << dp[1][n][(1 << n) - 1] << '\n';
    return 0;
}

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