理想的正方形

理想的正方形

二维倍增

SCUACM2022集训前训练-动态规划 - Virtual Judge (vjudge.net)

设 \(maxn[i][j][k]\) 为以 \((i,j)\) 为左上角，边长为 \(2^k\) 的正方形内元素的最大值

#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>

using namespace std;
typedef long long ll;
const int N = 1e3 + 10;
int n, m, len, Log[N];
int maxn[N][N][15], minn[N][N][15];
int a[N][N];
void init()
{
	for (int i = 2; i < N; i++)
		Log[i] = Log[i / 2] + 1;
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= m; j++)
			maxn[i][j][0] = minn[i][j][0] = a[i][j];
	for (int k = 1; k <= 12; k++)
	{
		for (int i = 1; i + (1 << k) - 1 <= n; i++)
		{
			for (int j = 1; j + (1 << k) - 1 <= m; j++)
			{
				int t1, t2, t3, t4;
				t1 = maxn[i][j][k-1];
				t2 = maxn[i + (1 << k - 1)][j][k-1];
				t3 = maxn[i][j + (1 << k - 1)][k-1];
				t4 = maxn[i + (1 << k - 1)][j + (1 << k - 1)][k-1];
				maxn[i][j][k] = max({t1, t2, t3, t4});
				
				t1 = minn[i][j][k-1];
				t2 = minn[i + (1 << k - 1)][j][k-1];
				t3 = minn[i][j + (1 << k - 1)][k-1];
				t4 = minn[i + (1 << k - 1)][j + (1 << k - 1)][k-1];
				minn[i][j][k] = min({t1, t2, t3, t4});
			}
		}
	}
}

int query(int sx, int sy, int len)
{
	int k = Log[len];
	int ex = sx + len - 1, ey = sy + len - 1;
	int t1, t2, t3, t4;
	t1 = maxn[sx][sy][k];
	t2 = maxn[sx][ey + 1 - (1 << k)][k];
	t3 = maxn[ex + 1 - (1 << k)][sy][k];
	t4 = maxn[ex + 1 - (1 << k)][ey + 1 - (1 << k)][k];
	int mx = max({t1, t2, t3, t4});
	
	t1 = minn[sx][sy][k];
	t2 = minn[sx][ey + 1 - (1 << k)][k];
	t3 = minn[ex + 1 - (1 << k)][sy][k];
	t4 = minn[ex + 1 - (1 << k)][ey + 1 - (1 << k)][k];
	int mn = min({t1, t2, t3, t4});
	// cout << sx << " " << sy << ": " << mx << " " << mn << endl;
	return mx - mn;
}

int main()
{
	scanf("%d%d%d", &n, &m, &len);
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= m; j++)
			scanf("%d", &a[i][j]);
	init();
	int ans = 2e9;
	for (int sx = 1; sx + len - 1 <= n; sx++)
		for (int sy = 1; sy + len - 1 <= m; sy++)
			ans = min(ans, query(sx, sy, len));
	printf("%d\n", ans);
	return 0;
}