제출 #503023

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
503023		JerryLiu06	구경하기 (JOI13_watching)	C++17	0 / 100	1 ms	332 KiB

watching

// https://oj.uz/problem/view/JOI13_watching

#include <bits/stdc++.h>
 
using namespace std;
 
using ll = long long;
using ld = long double;
using db = double;
using str = string;
 
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using pd = pair<db, db>;
 
using vi = vector<int>;
using vb = vector<bool>;
using vl = vector<ll>;
using vd = vector<db>;
using vs = vector<str>;
using vpi = vector<pi>;
using vpl = vector<pl>;
using vpd = vector<pd>;
 
#define FASTIO ios_base::sync_with_stdio(false); cin.tie(0);
#define FASTIOF ios_base::sync_with_stdio(false); fin.tie(0);
 
#define mp make_pair
#define f first
#define s second
 
#define sz(x) (int)(x).size()
#define bg(x) begin(x)
#define all(x) bg(x), end(x)
#define sor(x) sort(all(x))
#define rsz resize
#define ins insert 
#define ft front()
#define bk back()
#define pb push_back
#define pf push_front
 
#define lb lower_bound
#define ub upper_bound
 
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define F0R(i, a) FOR(i, 0, a)
#define ROF(i, a, b) for (int i = (b) - 1; i >= (a); i--)
#define R0F(i, a) ROF(i, 0, a)
#define EACH(a, x) for (auto& a : x)
 
ll cdiv(ll a, ll b) { return a / b + ((a ^ b) > 0 && a % b); }
ll fdiv(ll a, ll b) { return a / b - ((a ^ b) < 0 && a % b); }
 
template<class T> bool ckmin(T& a, const T& b) { return b < a ? a = b, 1 : 0; }
template<class T> bool ckmax(T& a, const T& b) { return a < b ? a = b, 1 : 0; }
 
template<class T> void remDup(vector<T>& v) { sor(v); v.erase(unique(all(v)), v.end()); }
 
const int MOD = 1e9;
const int MX = 2e3 + 10;
const ll INF = 1e18;

int N, P, Q, A[MX]; int nexLa[MX], nexSm[MX];

bool valid(int val) {
    int R1 = 1, R2 = 1;

    FOR(i, 1, N + 1) {
        while (R1 <= N && A[R1] - A[i] <= 2 * val - 1) {
            R1++;
        }
        nexLa[i] = R1 - 1;

        while (R2 <= N && A[R2] - A[i] <= val - 1) {
            R2++;
        }
        nexSm[i] = R2 - 1;
    }
    int DP[P + 1][Q + 1];

    F0R(i, P + 1) {
        F0R(j, Q + 1) {
            DP[i][j] = 0;

            if (i == 0 && j == 0) {
                continue;
            }
            if (i != 0) {
                ckmax(DP[i][j], nexSm[DP[i - 1][j] + 1]);
            }
            if (j != 0) {
                ckmax(DP[i][j], nexLa[DP[i][j - 1] + 1]);
            }
            if (DP[i][j] == N) {
                return true;
            }
        }
    }
    return false;
}

int main() {
    FASTIO;

    cin >> N >> P >> Q;

    ckmin(Q, N);
    ckmin(P, N - Q);

    FOR(i, 1, N + 1) {
        cin >> A[i];
    }
    int low = 1;
    int high = MOD;
    int res = -1;

    while (low <= high) {
        int mid = (low + high) / 2;

        if (valid(mid)) {
            res = mid, high = mid - 1;
        }
        else {
            low = mid + 1;
        }
    }
    cout << res;
}

• Subtask 1: 0 / 50
• Subtask 2: 0 / 50