Submission #618679

#	Time	Username	Problem	Language	Result	Execution time	Memory
618679		maomao90	Minerals (JOI19_minerals)	C++17	40 / 100	2841 ms	2804 KiB

minerals

// 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 "minerals.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 = 86005;

int n;
vi lft, rht;
bool state[MAXN];

void dnc(vi &fi, vi &se) {
    assert(SZ(fi) == SZ(se));
    cerr << "fi:";
    for (int i : fi) {
        cerr << ' ' << i;
    }
    cerr << '\n';
    cerr << "se:";
    for (int i : se) {
        cerr << ' ' << i;
    }
    cerr << '\n';
    if (SZ(fi) == 1) {
        Answer(fi[0], se[0]);
        return;
    }
    vi lfi, rfi;
    vector<bool> inlfi(2 * n + 1, 0);
    REP (i, 0, SZ(fi) / 2) {
        lfi.pb(fi[i]);
        inlfi[fi[i]] = 1;
    }
    REP (i, SZ(fi) / 2, SZ(se)) {
        rfi.pb(fi[i]);
    }
    int cur = -1;
    REP (i, 1, 2 * n + 1) {
        if (inlfi[i] ^ state[i]) {
            cur = Query(i);
            state[i] ^= 1;
        }
    }
    assert(cur != -1);
    vi lse, rse;
    for (int i : se) {
        int nxt = Query(i);
        if (nxt == cur) {
            lse.pb(i);
        } else {
            rse.pb(i);
        }
        cur = nxt;
    }
    dnc(lfi, lse);
    dnc(rfi, rse);
}

void Solve(int N) {
    n = N;
    int cur = 0;
    REP (i, 1, 2 * n + 1) {
        int nxt = Query(i);
        if (nxt > cur) {
            lft.pb(i);
        } else {
            rht.pb(i);
        }
        cur = nxt;
    }
    assert(SZ(lft) == n);
    REP (i, 1, 2 * n + 1) {
        Query(i);
    }
    dnc(lft, rht);
}

• Subtask 1: 6 / 6
• Subtask 2: 25 / 25
• Subtask 3: 9 / 9
• Subtask 4: 0 / 30
• Subtask 5: 0 / 5
• Subtask 6: 0 / 5
• Subtask 7: 0 / 5
• Subtask 8: 0 / 5
• Subtask 9: 0 / 10