#include <cstdio>
using namespace std;
int main(){
    int n,m,a[101][101];
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)scanf("%d",&a[i][j]);
    for(int i=2;i<n;i++)for(int j=2;j<m;j++)if(a[i][j]>a[i-1][j]&&a[i][j]>a[i+1][j]&&a[i][j]>a[i][j-1]&&a[i][j]>a[i][j+1])printf("%d\n",a[i][j]);
    return 0;
}

暴力枚举即可，时间复杂度O（n * m） AC代码！！！！！

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int m,n,a[105][105];
    cin >> n >> m;
    for (int i = 1;i <= n;i++)
    {
        for (int j = 1;j <= m;j++)
        {
            cin >> a[i][j];
        }
    }
    for (int i = 2;i < n;i++)
    {
        for (int j = 2;j < n;j++)
        {
            if (a[i][j] > a[i][j - 1] && a[i][j] > a[i][j + 1] && a[i][j] > a[i - 1][j] && a[i][j] > a[i + 1][j])
            {
                cout << a[i][j] << endl;
            }
        }
    }
}