# include <bits/stdc++.h>
using namespace std;
long long n, m;
long long f[1002][1002][2],a[1002][1002];
int main() {
	scanf("%lld%lld", &n, &m);
	for (long long i = 1; i <= n; i++) {
		for (long long j = 1; j <= m; j++) {
			scanf("%lld", &a[i][j]);
		}
	}
	memset(f, 128, sizeof(f));
	f[1][1][0] = a[1][1];
	for (long long i = 1; i <= m; i++) {
		for (long long j = 1; j <= n; j++) {
			if (i == 1 && j == 1) {
                continue;
            }
			f[j][i][0] = max(f[j][i - 1][0], max(f[j][i - 1][1], f[j - 1][i][0])) + a[j][i];
		}
		for (long long j = n; j >= 1; j--) {
			f[j][i][1] = max(f[j][i - 1][0], max(f[j][i - 1][1], f[j + 1][i][1])) + a[j][i];
		}
	}
	printf("%lld", f[n][m][0]);
	return 0;
}

本蒟蒻第一次写题解！ 如有不足之处请指出！

首先，这道题很明显是个动态规划，所以说，用动态规划暴力做一下就好了，以下为参考代码：（真的可以说非常暴力）

#include
using namespace std;
const int N=1007;
int a[N][N];
long long f[2][N][N];
//f[x][y][z] represents the point (y,z),which is the optimal solution in the Z direction.
//(when x=0 is transferred from top to bottom,when x=1 is transferred from bottom to top)
//f[0][i][j]=f[1][i][j]=max(f[0][i][j-1],f[1][i][j-1])+a[i][j];
int main(){
	int n,m;
	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=0;i<n+2;i++)for(int j=0;j<m+2;j++)f[0][i][j]=f[1][i][j]=-1e8*1ll;
	f[0][1][1]=f[1][1][1]=a[1][1];
	for(int i=2;i<=n;i++)f[0][i][1]=f[0][i-1][1]+a[i][1];
	for(int j=2;j<=m;j++){
		for(int i=1;i<=n;i++)f[0][i][j]=f[1][i][j]=max(f[0][i][j-1],f[1][i][j-1])+a[i][j];
		for(int i=2;i<=n;i++)f[0][i][j]=max(f[0][i][j],f[0][i-1][j]+a[i][j]);
		for(int i=n-1;i>=1;i--)f[1][i][j]=max(f[1][i][j],f[1][i+1][j]+a[i][j]);
	}
	printf("%lld",max(f[0][n][m],f[1][n][m]));
	return 0;
}

代码仅供参考！！切勿直接复制粘贴！

代码仅供参考！！切勿直接复制粘贴！

代码仅供参考！！切勿直接复制粘贴！

（重要的事情说三遍）

#include <stdio.h>
typedef long long LL;
const LL min_ll = -1e18;
int n, m; LL w[1005][1005], f[1005][1005][2];
inline LL mx(LL p, LL q, LL r) {return p > q ? (p > r ? p : r) : (q > r ? q : r);}
inline LL dfs(int x, int y, int from) {
    if (x < 1 || x > n || y < 1 || y > m) return min_ll;
    if (f[x][y][from] != min_ll) return f[x][y][from];
    if (from == 0) f[x][y][from] = mx(dfs(x + 1, y, 0), dfs(x, y - 1, 0), dfs(x, y - 1, 1)) + w[x][y];
    else f[x][y][from] = mx(dfs(x - 1, y, 1), dfs(x, y - 1, 0), dfs(x, y - 1, 1)) + w[x][y];
    return f[x][y][from];
}
int main(void) {
//	freopen("number.in", "r", stdin); freopen("number.out", "w", stdout);
	scanf("%d %d", &n, &m);
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= m; ++j) {
			scanf("%lld", &w[i][j]);
			f[i][j][0] = f[i][j][1] = min_ll;
		}
    f[1][1][0] = f[1][1][1] = w[1][1];
	printf("%lld\n", dfs(n, m, 1));
	return 0;
}