1 /*
 2     题意：抽象一点就是给两个矩阵，重叠的(就是两者选择其一)，两种铺路：从右到左和从下到上，中途不能转弯，
 3         到达边界后把沿途路上的权值相加求和使最大
 4     DP：这是道递推题，首先我题目看了老半天，看懂后写出前缀和又不知道该如何定义状态好，写不出状态转移方程，太弱了。
 5          dp[i][j]表示以(i, j)为右下角时求得的最大值，状态转移方程：dp[i][j] = max (dp[i-1][j] + sum1[i][j], dp[i][j-1] + sum2[i][j]);   sum1表示列的前缀，sum2表示行的前缀
 6 */
 7 /************************************************
 8 * Author        :Running_Time
 9 * Created Time  :2015-8-9 10:18:37
10 * File Name     :UVA_1366.cpp
11  ************************************************/
12
13 #include <cstdio>
14 #include <algorithm>
15 #include <iostream>
16 #include <sstream>
17 #include <cstring>
18 #include <cmath>
19 #include <string>
20 #include <vector>
21 #include <queue>
22 #include <deque>
23 #include <stack>
24 #include <list>
25 #include <map>
26 #include <set>
27 #include <bitset>
28 #include <cstdlib>
29 #include <ctime>
30 using namespace std;
31
32 #define lson l, mid, rt << 1
33 #define rson mid + 1, r, rt << 1 | 1
34 typedef long long ll;
35 const int MAXN = 5e2 + 10;
36 const int INF = 0x3f3f3f3f;
37 const int MOD = 1e9 + 7;
38 int a[MAXN][MAXN], b[MAXN][MAXN];
39 int sum1[MAXN][MAXN], sum2[MAXN][MAXN];
40 int dp[MAXN][MAXN];
41
42 int main(void)    {     //UVA 1366 Martian Mining
43     int n, m;
44     while (scanf ("%d%d", &n, &m) == 2) {
45         if (!n && !m)   break;
46         memset (sum1, 0, sizeof (sum1));
47         for (int i=1; i<=n; ++i)    {
48             for (int j=1; j<=m; ++j)    {
49                 scanf ("%d", &a[i][j]); sum1[i][j] = sum1[i][j-1] + a[i][j];
50             }
51         }
52         memset (sum2, 0, sizeof (sum2));
53         for (int i=1; i<=n; ++i)    {
54             for (int j=1; j<=m; ++j)    {
55                 scanf ("%d", &b[i][j]); sum2[i][j] = sum2[i-1][j] + b[i][j];
56             }
57         }
58         memset (dp, 0, sizeof (dp));
59         for (int i=1; i<=n; ++i)    {
60             for (int j=1; j<=m; ++j)    {
61                 dp[i][j] = max (dp[i-1][j] + sum1[i][j], dp[i][j-1] + sum2[i][j]);
62             }
63         }
64         printf ("%d\n", dp[n][m]);
65     }
66
67     return 0;
68 }