Submission #236172

#	Time	Username	Problem	Language	Result	Execution time	Memory
236172		ernestvw	Watching (JOI13_watching)	C++11	50 / 100	33 ms	16896 KiB

watching

/*
	* ernestvw
	* C++
	* May 2020
*/

#include <bits/stdc++.h>
using namespace std;

#define int long long

#define x first
#define y second
#define pb push_back
#define pii pair<int, int>
#define veci vector<int>
#define vecs vector<string>
#define amax(x, v) x = max(x, (v))
#define amin(x, v) x = min(x, (v))
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
#define sz(v) (int)(v.size())
#define FOR(i, n) for(int i = 0; i < (n); ++i)
#define nl '\n'

using ll = long long;
using ld = long double;
using ull = unsigned long long;
template <class T>
using PQ = priority_queue<T, vector<T>, greater<T>>;

const int oo = 1e9; // INT_MAX
const long long OO = 1e18; // LLONG_MAX
const double eps = 1e-9;
const double pi = 4.0 * atan(1.0);
const int p197 = 1000000007;
const int p998 = 998244353;
const int p103 = 1000003;

#define TEST_CASES 0

void solve(int test_case);

signed main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL); cout.tie(NULL);

	int T = 1;
	if(TEST_CASES) cin >> T;
	for(int t = 1; t <= T; t++)
		solve(t);

	return 0;
}

// ---------------------------------------------------

int n, p, q;
veci a;

int dp[101][101][101];
int W = 1;

int f(int i, int P, int Q) {
	if(i < n && P == 0 && Q == 0) {
		return 0;
	}
	if(i == n) return 1;
	if(dp[i][P][Q] != -1) return dp[i][P][Q];
	dp[i][P][Q] = 0;
	if(P > 0) {
		int j = i;
		while(j < n && a[j] - a[i] + 1LL <= W) j++;
		dp[i][P][Q] |= f(j, P-1, Q);
	}
	if(Q > 0) {
		int j = i;
		while(j < n && a[j] - a[i] + 1LL <= 2LL * W) j++;
		dp[i][P][Q] |= f(j, P, Q-1);
	}
	return dp[i][P][Q];
}

bool possible(int w) {
	W = w;
	memset(dp, -1, sizeof(dp));
	return bool(f(0, p, q));
}

void solve(int test_case) {
	cin >> n >> p >> q;
	a.resize(n);
	FOR(i, n) cin >> a[i];
	sort(all(a));
	amin(p, n), amin(q, n);
	int w = 1e9;
	for(int j = 31; j >= 0; j--) {
		if(w - (1LL << j) >= 0 && possible(w - (1LL << j))) {
			w -= (1LL << j);
		}
	}
	cout << w << endl;
}

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