我们需要先从数组较大的开始进行处理，每次考察上下左右的，比较当前存储的最大值和转移来的值，哪一个大一点

#define _CRT_SECURE_NO_WARNINGS
#include<bits/stdc++.h>
using namespace std;

int n, m;
int a[105][105];
int addx[] = { 0,0,1,-1 };
int addy[] = { 1,-1,0,0 };

struct node
{
	int va;
	int x, y;
	bool operator<(node b) {
		return va > b.va;
	}
}sto[10010];

int dp[105][105];

int main() {
	cin >> n >> m;
	int cnt = 0;
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= m; j++) {
			cin >> a[i][j];
			sto[cnt].va = a[i][j];
			sto[cnt].x = i;
			sto[cnt].y = j; cnt++;
		}
	}

	
	sort(sto, sto + cnt);
	for (int i = 0; i < cnt; i++) {
		int now = sto[i].va;
		int x = sto[i].x;
		int y = sto[i].y;
		for (int i = 0; i < 4; i++) {
			int xx = x + addx[i];
			int yy = y + addy[i];
			if (xx<0 || yy <0 || xx>n || yy>m) continue;
			if (now > a[xx][yy]) dp[xx][yy] = max(dp[xx][yy],dp[x][y]+1);
		}
	}
	int ans = 0;

	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= m; j++) {
			ans = max(ans, dp[i][j]);
		}
	}
	cout << ans+1;
	return 0;
}