摘花生 https://www.acwing.com/problem/content/1017/

从网格状左上角起点，只能向下和向右走，走到右下角终点
每个点有对应权值，问走完后，经过点的权值之和最大为多少

动态规划 1）状态表示 1）集合：f[i][j]表示走到(i,j)的所有走法
                  2）属性：Max
        2）状态计算 

状态转移方程 f[i][j] = max(f[i][j - 1] + f[i][j], f[i - 1][j] + f[i][j])

#include <iostream>
using namespace std;

const int N = 105;
int t, f[N][N];

int main()
{
    scanf("%d", &t);
    while ( t -- )
    {
        int n, m;
        scanf("%d%d", &n, &m);
        for ( int i = 1; i <= n; i ++ )
            for ( int j = 1; j <= m; j ++ )
            {
                cin >> f[i][j];
                f[i][j] = max(f[i - 1][j] + f[i][j], f[i][j - 1] + f[i][j]);
            }
        printf("%d\n", f[n][m]);
    }
    return 0;
}