제출 #236175

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
236175		ernestvw	구경하기 (JOI13_watching)	C++11	100 / 100	98 ms	12024 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[2001][2001];
int prochain[2001], prochain2[2001];

bool possible(int w) {
	int p1 = 0, p2 = 0;
	FOR(i, n) {
		while(p1 < n && a[p1] - a[i] + 1LL <= w) p1++;
		while(p2 < n && a[p2] - a[i] + 1LL <= 2LL * w) p2++;
		prochain[i] = p1;
		prochain2[i] = p2;
	}
	FOR(i,p+1)FOR(j,q+1){
		if(i==0 && j==0){
			dp[i][j]=0;
			continue;
		}
		dp[i][j] = 0;
		if(i > 0) {
			if(dp[i-1][j] == n) dp[i][j] = n;
			else {
				amax(dp[i][j], prochain[dp[i-1][j]]);
			}
		}
		if(j > 0) {
			if(dp[i][j-1] == n) dp[i][j] = n;
			else {
				amax(dp[i][j], prochain2[dp[i][j-1]]);
			}
		}
	}
	return dp[p][q] >= n;
}

void solve(int test_case) {
	cin >> n >> p >> q;
	a.resize(n);
	FOR(i, n) cin >> a[i];
	sort(all(a));
	if(p + q >= n) {
		cout << 1 << endl;
		return;
	}
	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: 50 / 50