Problem Description

Given a matrix of n rows and m columns,find the largest area submatrix which is non decreasing on each column

Input

The first line contains an integer T(1≤T≤10)representing the number of test cases.
For each test case, the first line contains two integers n,m(1≤n,m≤2∗103)representing the size of the matrix
the next n line followed. the i-th line contains m integers vij(1≤vij≤5∗103)representing the value of matrix
It is guaranteed that there are no more than 2 testcases with n∗m>10000

Output

For each test case, print a integer representing the Maximal submatrix

Sample Input

1
2 3
1 2 4
2 3 3

Sample Output

4

\(n^2\)DP，设dp[i, j]为mtx[i, j]为右下角的最大的满足题意的子矩阵大小。按行遍历再按列遍历，对于(i, j)这个位置，首先求出来这一列以(i, j)结尾的最长单调不减序列的长度（维护一个数组mx即可）。然后再重新遍历第i行对mx数组跑单调栈更新dp[i, j]即可。

比如对于:

1 2 2 3
2 1 3 4
5 2 2 4

第三行的mx数组为[3, 2, 1, 3]，则dp[3, 1] = 3，dp[3, 2] = 4， dp[3, 3] = 3，dp[3, 4] = 4。

#include <bits/stdc++.h>
using namespace std;
#define int long long
int mtx[2005][2005], n, m;

int dp[2005][2005], mx[2005], s[2005], w[2005];
signed main() {
	ios::sync_with_stdio(false);
	int t;
	cin >> t;
	while(t--) {
		cin >> n >> m;
		memset(dp, 0, sizeof(dp));
		memset(mx, 0, sizeof(mx));
		for(int i = 1; i <= n; i++) {
			for(int j = 1; j <= m; j++) {
				dp[i][j] = 1;
			}
		}
		for(int i = 1; i <= n; i++) {
			for(int j = 1; j <= m; j++) {
				cin >> mtx[i][j];
			}
		}
		int ans = 0;
		for(int i = 1; i <= n; i++) {
			for(int j = 1; j <= m; j++) {//dp[i][j]表示i, j结尾的最大的满足条件的子矩阵
				if(mtx[i][j] >= mtx[i - 1][j]) {
					mx[j]++;
				} else {
					mx[j] = 1;
				}
			}
			//mx[1 ~ j]从中选一个连续子区间l, r,设区间最小值为mn，怎么选使得mn * (r - l + 1）最大
			//单调栈
			int p;
			mx[m + 1] = p = 0;
			s[0] = s[1] = 0;
			w[0] = w[1] = 0;
			for(int j = 1; j <= m + 1; j++) {
				if(mx[j] > s[p]) {
					s[++p] = mx[j];
					w[p] = 1;
				} else {
					int width = 0;
					while(s[p] > mx[j]) {
						width += w[p];
						dp[i][j] = max(dp[i][j], width * s[p]);
						ans = max(ans, dp[i][j]);
						p--;
					}
					s[++p] = mx[j], w[p] = width + 1;
				}
			}
		}
		cout << ans << endl;
	}
	return 0;
}