水题，记忆化搜索，队列bfs均可

我们定义f[i][j]为到(i, j)的最长路径。然后就不难得出状态转移方程，然后使用无脑dfs，或者有脑递推都是可以的。

#include <bits/stdc++.h>

using namespace std;

const int maxn = 105;

const int dx[] = {0, 1, 0, -1};

const int dy[] = {1, 0, -1, 0};

struct node

{

    int x, y;

};

int anss = 0;

int table[maxn][maxn];

int hi = 0;

int n, m;

queue<node> q;

int dp[maxn][maxn];

int dfs(int x, int y) {

    int maxx = 0;

    if(x == 0 || y == 0) return dp[x][y] = 0;

    if(dp[x][y]) return dp[x][y];

    for(int i = 0; i < 4; i++) {

        int nx = x + dx[i];

        int ny = y + dy[i];

        if(table[nx][ny] > table[x][y]) maxx = max(maxx, dfs(nx, ny));

    }

    return dp[x][y] = maxx + 1;

}

int main() {

    cin >> n >> m;

    for(int i = 1; i <= n; i++) {

        for(int j = 1; j <= m; j++) {

            cin >> table[i][j];

            if(table[i][j] > hi) {

                while(!q.empty()) q.pop();

                q.push((node){i, j});

                hi = table[i][j];

            }

            if(table[i][j] == hi) {

                q.push((node){i, j});

            }

        }

    }

    while(!q.empty()) {

        node fr = q.front(); q.pop();

        dp[fr.x][fr.y] = 1;

        for(int i = 1; i <= n; i++) {

            for(int j = 1; j <= m; j++) {

                anss = max(anss, dfs(i, j));

            }

        }

    }

    cout << anss;

}