xsy子矩形

• 考虑一种解题方法，枚举上下边界L,R， 然后二分答案T，我们要判断的是否存在

\[\frac{(sum_j - sum_i)}{2 * (R - L + 1 + j - i)} \ge T
\]

• 也就是

\[(sum_j - 2Tj) - (sumi - 2Ti) \ge 2T(R - L + 1)
\]

• 维护前缀最小值即可
• 然后这样是三次方的， 我们考虑random_shuffle一下枚举的序列， 然后查询的时候就先看一下当前的ans + eps是否可行， 不可行的话就不进行二分直接退出
• 这样期望是会进行lnn次的二分， 就赢了

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<iostream>
#define M 515
#define ll long long
#define mmp make_pair
using namespace std;
int read() {
	int nm = 0, f = 1;
	char c = getchar();
	for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
	return nm * f;
}
const double eps = 1e-10, inf = 5e13;
double ans = -1e9;
ll sum[M][M], n, tp, ver[M], xl, xr, yl, yr;
pair<int, int> note[M * M];
ll get(int xa, int ya, int xb, int yb) {
	return sum[xb][yb] - sum[xa - 1][yb] - sum[xb][ya - 1] + sum[xa - 1][ya - 1];
}

bool check(double T, int len, int l, int r) {
	double minn = 0, pl = 0;
	double up = T * 2 * len;
	for(int i = 1; i <= n; i++) {
		double ver2 = ver[i] - T * 2 * i;
		if(ver2 - minn >= up) {
			if(T > ans) {
				ans = T;
				xl = l, yl = pl + 1, xr = r, yr = i;
			}
			return true;
		}
		if(ver2 < minn) minn = ver2, pl = i;
	}
	return false;
}

int main() {
	//freopen("rectangle3.in", "r", stdin);
	n = read();
	for(int i = 1; i <= n; i++) for(int j = 1; j <= n; j++) sum[i][j] = sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1] + read();
	for(int i = 1; i <= n; i++) for(int j = i; j <= n; j++) note[++tp] = mmp(i, j);
	for(int i = 1; i <= tp; i++) {
		int x = 1ll * rand() * rand() % i + 1;
		swap(note[x], note[i]);
	}
	for(int i = 1; i <= tp; i++) {
		int l = note[i].first, r = note[i].second;
		for(int j = 1; j <= n; j++) ver[j] = get(l, 1, r, j);
		if(!check(ans + eps, r - l + 1, l, r)) continue;
		double L = ans, R = inf;
		for(int j = 1; j <= 100; j++) {
			double mid = (L + R) / 2.0;
			if(!check(mid, r - l + 1, l, r)) R = mid;
			else L = mid;
		}
	}
//	cerr<< note[tp].first << " " << note[tp].second << "\n";
	printf("%.10lf\n", ans);
	cout << xl << " " << yl << "\n" << xr << " " << yr << "\n";
	return 0;
}