题目来源：acwing 275 传纸条 

分析：这题和两人同时摘樱桃之类的题一样，一个人从左上角走到右下角，再从右下角走回左上角，相同地点的分数只能得一次（或者不能走相同地点）。这种题统一可以按照两个人从左上角一起出发，走到右下角，相同格子得分只能算一次来计算。

状态定义：f[k][i1][i2] 代表从(0,0)出发，已经走了k步，两个人横坐标分别为i1,i2时，得分最大值

结果：f[m + n][m + 1][m + 1] 从0走到m一共走m步，从(0,0)走到(m,n) 一共要走m + n步，横坐标范围从0-m共m+1

状态转移：f[k][i1][i2] 从f[k-1]转移过来，(i,j) 从(i-1,j),(i,j - 1)转移过来，当i1 = i2时，j1==j2，同一个点注意只能取一次，这种情况肯定小于两个点不同的情况，如果分数有负数，这里可以考虑加上一个负无穷，所以最终答案肯定是两个人走的路径不同的更优情况。

m,n = map(int,input().split())
g = []
for _ in range(m):g.append(list(map(int,input().split())))f = [[[0]*(m + 1) for _ in range(m + 1)] for _ in range(m + n + 1)]
for k in range(2,m + n + 1):for i1 in range(1,m + 1):for i2 in range(1,m + 1):j1,j2 = k - i1,k - i2if 1 <= j1 <= n and 1 <= j2 <= n:x = g[i1 - 1][j1 - 1] + (g[i2 - 1][j2 - 1] if i1 != i2 else 0)f[k][i1][i2] = max(f[k - 1][i1 - 1][i2 - 1],f[k - 1][i1][i2],f[k - 1][i1 - 1][i2],f[k - 1][i1][i2 - 1]) + x
print(f[-1][-1][-1])

类似题目：https://leetcode.cn/problems/cherry-pickup/

class Solution:def cherryPickup(self, grid: List[List[int]]) -> int:n = len(grid)if grid[0][0] == -1:return 0# 两人一起走f = [[[-inf]*(n +1) for _ in range(n + 1)] for _ in range(n + n + 1)]f[2][1][1] = grid[0][0]for k in range(2,n + n + 1):for i1 in range(1,n + 1):for i2 in range(1,n + 1):if k == 2 and i1 == 1 and i2 == 1:continuej1,j2 = k - i1,k - i2if 1 <= j1 <= n and 1 <= j2 <= n and grid[i1 - 1][j1 - 1] >= 0 and grid[i2 - 1][j2 - 1] >= 0:x = grid[i1 - 1][j1 - 1] + (grid[i2 - 1][j2 - 1] if i1 != i2 else 0)f[k][i1][i2] = max(f[k - 1][i1 - 1][i2 - 1],f[k - 1][i1][i2],f[k - 1][i1 - 1][i2],f[k - 1][i1][i2 - 1]) + xreturn max(f[-1][-1][-1],0)

https://leetcode.cn/problems/cherry-pickup-ii/

class Solution:def cherryPickup(self, grid: List[List[int]]) -> int:m,n = len(grid),len(grid[0])f = [[[-inf]*n for _ in range(n)] for _ in range(m)]f[0][0][-1] = grid[0][0] + grid[0][-1]for i in range(1,m):for j1 in range(n):for j2 in range(n):x = grid[i][j1] + (grid[i][j2] if j1 != j2 else 0)for y1 in (j1 - 1,j1,j1 + 1):for y2 in (j2 - 1,j2,j2 + 1):if 0 <= y1 < n and 0 <= y2 < n:f[i][j1][j2] = max(f[i - 1][y1][y2] + x,f[i][j1][j2])return max(f[-1][j][j1] for j in range(n) for j1 in range(n))