Submission #362065

#	Time	Username	Problem	Language	Result	Execution time	Memory
362065		Galebickosikasa	The Kingdom of JOIOI (JOI17_joioi)	C++17	100 / 100	625 ms	80620 KiB

joioi

// #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,sse4.1,sse4.2,popcnt,abm,mmx,avx")
 
#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
#include <unordered_map>
#include <set>
#include <map>
#include <queue>
#include <random>
#include <chrono>

#define fi first
#define se second
#define pb push_back
#define ll long long
#define ld long double
#define hm unordered_map 
#define pii pair<int, int>
#define sz(a) (int)a.size()
#define all(a) a.begin(), a.end()
#define cinv(v) for (auto& x: v) cin >> x
#define fr(i, n) for (int i = 0; i < n; ++i)
#define fl(i, l, n) for (int i = l; i < n; ++i)

// #define int ll

template <typename T1, typename T2> inline bool chkmin(T1 &x, const T2 &y) {if (x > y) {x = y; return 1;} return 0;}
template <typename T1, typename T2> inline bool chkmax(T1 &x, const T2 &y) {if (x < y) {x = y; return 1;} return 0;}

using namespace std;

#ifdef LOCAL
	#define dbg(x) cerr << #x << " : " << x << '\n'
#else 
	#define dbg(x)
#endif

//tg: @runningcherry
 
template <typename T1, typename T2> ostream& operator << (ostream& out, const pair<T1, T2>& v) {
	out << v.fi << ", " << v.se;
	return out;
}

template<typename T> ostream& operator << (ostream& out, const vector<T>& v) {
	for (auto& x: v) out << x << " ";
	return out;
}

template <typename T1, typename T2> istream& operator >> (istream& in, pair<T1, T2>& a) {
	in >> a.fi >> a.se;
	return in;
}

const ll inf = (ll) 1e9;
const ld pi = asin (1) * 2;
const ld eps = 1e-8;
const ll mod = (ll)1e9 + 7;
const ll ns = 97;

const int maxn = 2000 + 20;

mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());

int goo[maxn][maxn], n, m, mxp[maxn][maxn], mnp[maxn][maxn], mxs[maxn][maxn], mns[maxn][maxn], mn = inf;
set <int> gay;

inline int checkl (int x) {
	vector<int> kek (n, -1);
	int mx = mn + x;
	fr (i, n) {
		while (kek[i] != -1 && mnp[i][kek[i]] < mn) --kek[i];
		while (kek[i] + 1 < m && mnp[i][kek[i] + 1] >= mn && mxp[i][kek[i] + 1] <= mx) ++kek[i];
	}
	int last = kek[0];
	int mx1 = -inf, mn1 = inf;
	fr (i, n) {
		chkmin (last, kek[i]);
		if (last + 1 < m) {
			chkmin (mn1, mns[i][last + 1]);
			chkmax (mx1, mxs[i][last + 1]);
		}
	}
	if (mx1 - mn1 <= x) return 1;
	last = kek.back ();
	mx1 = -inf, mn1 = inf;
	for (int i = n - 1; i >= 0; --i) {
		chkmin (last, kek[i]);
		if (last + 1 < m) {
			chkmin (mn1, mns[i][last + 1]);
			chkmax (mx1, mxs[i][last + 1]);
		}
	}
	if (mx1 - mn1 <= x) return 1;
	return 0;
}

inline int checkr (int x) {
	vector<int> kek (n, m);
	int mx = mn + x;
	fr (i, n) {
		while (kek[i] != m && mns[i][kek[i]] < mn) ++kek[i];
		while (kek[i] > 0 && mns[i][kek[i] - 1] >= mn && mxs[i][kek[i] - 1] <= mx) --kek[i];
	}
	int last = kek[0];
	int mx1 = -inf, mn1 = inf;
	fr (i, n) {
		chkmax (last, kek[i]);
		if (last > 0) {
			chkmin (mn1, mnp[i][last - 1]);
			chkmax (mx1, mxp[i][last - 1]);
		}
	}
	if (mx1 - mn1 <= x) return 1;
	last = kek.back ();
	mx1 = -inf, mn1 = inf;
	for (int i = n - 1; i >= 0; --i) {
		chkmax (last, kek[i]);
		if (last > 0) {
			chkmin (mn1, mnp[i][last - 1]);
			chkmax (mx1, mxp[i][last - 1]);
		}
	}
	if (mx1 - mn1 <= x) return 1;
	return 0;
}

int bs () {
	int l = -1, r = 1e9;
	while (r - l > 1) {
		int m = (r + l) >> 1;
		if (checkl (m) || checkr (m)) r = m;
		else l = m;
	}
	return r;
}

signed main () {
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);
	cin >> n >> m;
	fr (i, n) fr (j, m) cin >> goo[i][j];
	fr (i, n) {
		mnp[i][0] = mxp[i][0] = goo[i][0];
		fl (j, 1, m) {
			mnp[i][j] = min (mnp[i][j - 1], goo[i][j]);
			mxp[i][j] = max (mxp[i][j - 1], goo[i][j]);
		}
		mns[i][m - 1] = mxs[i][m - 1] = goo[i][m - 1];
		for (int j = m - 2; j >= 0; --j) {
			mns[i][j] = min (mns[i][j + 1], goo[i][j]);
			mxs[i][j] = max (mxs[i][j + 1], goo[i][j]);
		}
	}
	fr (i, n) fr (j, m) chkmin (mn, goo[i][j]);
	cout << bs () << '\n';






}

• Subtask 1: 15 / 15
• Subtask 2: 45 / 45
• Subtask 3: 40 / 40