使用动态规划求解，思想还是很简单的，lahub只能向右和向下，lahubina只能向右和向上，限制条件就是只能相遇一次。

分析一下就能知道除边界点外其他所有点都有可能成为相遇点，对相遇点进行分析，要使满足只相遇一次的条件，只能有以下两种情况：lahub向下走到相遇点并继续向下走，此时lahubina向右走到相遇点并继续向右走；另外一种情况就是lahub向右走到相遇点并继续向右走，lahubina向上走到相遇点并继续向上走。问题就解决了~

还有就是自己处理数组的时候一直喜欢下标从0开始计算，其实在很多时候，尤其是动规、贪心等算法里将数组的行列比实际大小都分大1或者2在求解的时候是能带来很大便利的，记住啦~

下面就是这道题的核心代码：

We can precalculate for dynamic programming matrixes and we're done.

dp1[i][j] = maximal cost of a path going from (1, 1) to (i, j) only down and right.

dp2[i][j] = maximal cost of a path from (i, j) to (1, m) going only up and right.

dp3[i][j] = maximal cost of a path from (m, 1) to (i, j) going only up and right.

dp4[i][j] = maximal cost of a path from (i, j) to (n, m) going only down or right.

And here is my full implementation of recurrences (C++ only):

for (int i = 1; i <= n; ++i)for (int j = 1; j <= m; ++j)dp1[i][j] = a[i][j] + max(dp1[i - 1][j], dp1[i][j - 1]);
for (int j = m; j >= 1; --j)for (int i = 1; i <= n; ++i)dp2[i][j] = a[i][j] + max(dp2[i - 1][j], dp2[i][j + 1]);
for (int i = n; i >= 1; --i)for (int j = 1; j <= m; ++j)dp3[i][j] = a[i][j] + max(dp3[i + 1][j], dp3[i][j - 1]);
for (int i = n; i >= 1; --i)for (int j = m; j >= 1; --j)dp4[i][j] = a[i][j] + max(dp4[i][j + 1], dp4[i + 1][j]);