ABC 203 D - Pond

D - Pond

二分 + 前缀和

求一个子矩阵中元素的中位数，可二分中位数，check 过程中将矩阵变为 01 矩阵，若小于 mid 则为 1，用二维前缀和求出有多少元素小于 mid

复杂度 \(O(n^2log(max(A_i)))\)

注意二分求中位数的细节

1. 若有 \(n\) 个数，则中位数为第 \(\lceil \frac n2\rceil\) 小的元素
2. 令 \(t=\lceil \frac n2\rceil-1\), 若有 \(cnt\) 个元素小于 mid, 返回 \(cnt<=t\)

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

using namespace std;
typedef long long ll;
const int N = 8e2 + 10;
int n, k, t;
int a[N][N], s[N][N];

bool check(int x)
{
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= n; j++)
			s[i][j] = s[i-1][j] + s[i][j-1] - s[i-1][j-1] + (a[i][j] < x);
	int cnt = 0;
	for (int sx = 1; sx + k - 1 <= n; sx++)
	{
		for (int sy = 1; sy + k - 1 <= n; sy++)
		{
			int ex = sx + k - 1, ey = sy + k - 1;
			cnt = max(cnt, s[ex][ey] - s[sx-1][ey] - s[ex][sy-1] + s[sx-1][sy-1]);
		}
	}
	return cnt <= t;
}
int main()
{
	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cin >> n >> k;
	
	t = (k * k + 1) / 2 - 1;//小于中位数的有t个
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= n; j++)
			cin >> a[i][j];
	int l = 0, r = 1e9 + 1;
	while(l + 1 != r)
	{
		int mid = l + r >> 1;
		if (check(mid))
			l = mid;
		else
			r = mid;
	}
	cout << l << endl;
	return 0;
}