Submission #211249

#	Time	Username	Problem	Language	Result	Execution time	Memory
211249		mode149256	Watching (JOI13_watching)	C++14	0 / 100	13 ms	384 KiB

watching

/*input
13 3 2
33
66
99
10
83
68
19
83
93
53
15
66
75

3 1 1
2
11
17
*/
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<ld, ld> pd;

typedef vector<int> vi;
typedef vector<vi> vii;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<vl> vll;
typedef vector<pi> vpi;
typedef vector<vpi> vpii;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
typedef vector<pd> vpd;
typedef vector<bool> vb;
typedef vector<vb> vbb;
typedef std::string str;
typedef std::vector<str> vs;

#define x first
#define y second
#define debug(...) cout<<"["<<#__VA_ARGS__<<": "<<__VA_ARGS__<<"]\n"

const int MOD = 1000000007;
const ll INF = std::numeric_limits<ll>::max();
const int MX = 100101;
const ld PI = 3.14159265358979323846264338327950288419716939937510582097494L;

template<typename T>
pair<T, T> operator+(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x + b.x, a.y + b.y); }
template<typename T>
pair<T, T> operator-(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x - b.x, a.y - b.y); }
template<typename T>
T operator*(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.x + a.y * b.y); }
template<typename T>
T operator^(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.y - a.y * b.x); }

template<typename T>
void print(vector<T> vec, string name = "") {
	cout << name;
	for (auto u : vec)
		cout << u << ' ';
	cout << '\n';
}

int N, P, Q;
vi sk;

bool galima(int m) {
	if (P + Q >= N) return true;
	vi dp(N + 1);
	for (int i = 0; i <= N; ++i)
		dp[i] = i;

	// if (m == 9) {
	// 	for (int i = 0; i <= N; ++i)
	// 		printf("%2d ", dp[i]);
	// 	printf("\n");
	// }

	for (int p = 0; p < P; ++p)
	{
		int j = N;
		for (int n = N; n >= 1; --n)
		{
			while (j and sk[n] - sk[j] + 1 <= m)
				j--;

			dp[n] = dp[j];
		}
		// if (m == 9) {
		// 	for (int i = 0; i <= N; ++i)
		// 		printf("%2d ", dp[i]);
		// 	printf("\n");
		// }
	}

	for (int p = 0; p < Q; ++p)
	{
		int j = N;
		for (int n = N; n >= 1; --n)
		{
			while (j and sk[n] - sk[j] + 1 <= 2 * m)
				j--;

			dp[n] = dp[j];
		}
		// if (m == 9) {
		// 	for (int i = 0; i <= N; ++i)
		// 		printf("%2d ", dp[i]);
		// 	printf("\n");
		// }
	}

	return !dp[N];
}

int main() {
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	cin >> N >> P >> Q;
	sk = vi(N + 1);
	for (int i = 1; i <= N; ++i)
		cin >> sk[i];

	sort(sk.begin() + 1, sk.end());

	// for (int i = 0; i <= N; ++i)
	// 	printf("%2d ", sk[i]);
	// printf("\n");
	int l = 1;
	int h = (int)1e9;
	int m;
	while (l < h) {
		m = (l + h) / 2;
		if (galima(m))
			h = m;
		else
			l = m + 1;
	}

	printf("%d\n", l);
}

/* Look for:
* special cases (n=1?)
* overflow (ll vs int?)
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* array bounds
*/

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