Submission #580612

#	Time	Username	Problem	Language	Result	Execution time	Memory
580612		maomao90	On the Grid (FXCUP4_grid)	C++17	100 / 100	21 ms	344 KiB

grid

// 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> 
#include "grid.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 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 = 1005;

int n;

vi shift(vi p) {
    vi res = p;
    REP (i, 0, SZ(p)) {
        if (i - 1 < 0) {
            res[i] = p[i - 1 + SZ(p)];
        } else {
            res[i] = p[i - 1];
        }
    }
    return res;
}

vi SortDisks(int N) {
    n = N;
    vi ans(n, -1), mpa(n + 1, -1);
    vi p(n, 0);
    iota(ALL(p), 0);
    int prv = PutDisks(p);
    while (1) {
        vi unfix;
        REP (i, 0, n) {
            if (ans[p[i]] == -1) {
                unfix.pb(p[i]);
            }
        }
        if (unfix.empty()) {
            break;
        }
        unfix = shift(unfix);
        reverse(ALL(unfix));
        vi np;
        REP (i, 0, n) {
            if (mpa[i + 1] != -1) {
                np.pb(mpa[i + 1]);
            } else {
                np.pb(unfix.back()); unfix.pop_back();
            }
        }
        p = np;
        int cur = PutDisks(p);
        if (cur >= prv) {
            ans[p[0]] = cur - (n - 1);
            mpa[cur - (n - 1)] = p[0];
        } else {
            prv = cur;
        }
    }
    return ans;
}

• Subtask 1: 12 / 12
• Subtask 2: 31 / 31
• Subtask 3: 57 / 57