简单DP
注意最后取值即可
//#pragma GCC optimize(2)
#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <cmath>
#include <cctype>
#include <string>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <ctime>
#include <vector>
#include <fstream>
#include <list>
#include <iomanip>
#include <numeric>
using namespace std;
#define int long long
const int MAXN = 2e3 + 10;
int arr[MAXN][MAXN];
int dp[MAXN][MAXN];
int n, m;
signed main()
{
//ios::sync_with_stdio(false);
//cin.tie(0); cout.tie(0);
//freopen("D://test.in", "r", stdin);
//freopen("D://test.out", "w", stdout);
cin>>n>>m;
int mid = m / 2 + 1;
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++)
cin>>arr[i][j];
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++)
dp[i][j] = max(max(dp[i-1][j], dp[i-1][j-1]), dp[i-1][j+1]) + arr[i][j];
int ans = -0x3f3f3f3f;
ans = max(max(dp[n][mid-1],dp[n][mid]),dp[n][mid+1]);
cout<<ans<<endl;
return 0;
}