题目：
Given a rows x cols matrix grid representing a field of cherries. Each cell in grid represents the number of cherries that you can collect.
You have two robots that can collect cherries for you, Robot #1 is located at the top-left corner (0,0) , and Robot #2 is located at the top-right corner (0, cols-1) of the grid.
Return the maximum number of cherries collection using both robots by following the rules below:
- From a cell (i,j), robots can move to cell (i+1, j-1) , (i+1, j) or (i+1, j+1).
- When any robot is passing through a cell, It picks it up all cherries, and the cell becomes an empty cell (0).
- When both robots stay on the same cell, only one of them takes the cherries.
- Both robots cannot move outside of the grid at any moment.
- Both robots should reach the bottom row in the
grid.
Example 1:
Input: grid = [[3,1,1],[2,5,1],[1,5,5],[2,1,1]]
Output: 24
Explanation: Path of robot #1 and #2 are described in color green and blue respectively.
Cherries taken by Robot #1, (3 + 2 + 5 + 2) = 12.
Cherries taken by Robot #2, (1 + 5 + 5 + 1) = 12.
Total of cherries: 12 + 12 = 24.
Example 2:
Input: grid = [[1,0,0,0,0,0,1],[2,0,0,0,0,3,0],[2,0,9,0,0,0,0],[0,3,0,5,4,0,0],[1,0,2,3,0,0,6]]
Output: 28
Explanation: Path of robot #1 and #2 are described in color green and blue respectively.
Cherries taken by Robot #1, (1 + 9 + 5 + 2) = 17.
Cherries taken by Robot #2, (1 + 3 + 4 + 3) = 11.
Total of cherries: 17 + 11 = 28.
Example 3:
Input: grid = [[1,0,0,3],[0,0,0,3],[0,0,3,3],[9,0,3,3]]
Output: 22
Example 4:
Input: grid = [[1,1],[1,1]]
Output: 4
Constraints:
rows == grid.lengthcols == grid[i].length2 <= rows, cols <= 700 <= grid[i][j] <= 100
思路：
三维DP，f[i][j][k]代表了第i行，robot1在第j列，robot2在第k列可以获得的最大的樱桃数。对于初始化，显然第一行是固定的，即f[0][0][n-1]应该等于grid[0][0]+grid[0][n-1]，n是列数。我们不用在乎f[0][0][0]，因为这种情况并不会发生。这里比较巧妙的地方是用当前的列去判断，如果按照一般思维，我们会在第i行判断i+1行的三个能否取到，有没有超出边界；但是这里因为第0行是固定的，因此我们可以从第1行开始循环，我们对于j和k，给定6个数，即j-1,j,j+1和k,k-1,k+1，只要i-1行的这6个中，各自的三个没有全超出边界，那么就可以取到[i][j][k]这一格。然后DP逻辑的话是如果j和k重叠，那么只要计算上一行的这六个格子+grid[i][j]和原本f[i][j][k]的最大值即可；如果不重叠，那么前面就要再加上gird[i][k]，因为两个机器人各自占了不同格子。
代码：
class Solution {
public:
int cherryPickup(vector<vector<int>>& grid) {
int m=grid.size(), n=grid[0].size();
vector<vector<vector<int>>> f(m,vector<vector<int>>(n,vector<int>(n,INT_MIN)));
f[0][0][n-1]=grid[0][0]+grid[0][n-1];
int count =0;
for(int i=1;i<m;i++)
{
for(int j=0;j<n;j++)
{
for(int offset1=-1;offset1<=1;offset1++)
{
int y1=j+offset1;
if(y1<0||y1>=n)
continue;
for(int k=0;k<n;k++)
{
for(int offset2=-1;offset2<=1;offset2++)
{
int y2=k+offset2;
if(y2<0|y2>=n)
continue;
if(j==k)
f[i][j][k]=max(f[i][j][k], f[i-1][y1][y2]+grid[i][j]);
else
f[i][j][k]=max(f[i][j][k], f[i-1][y1][y2]+grid[i][j]+grid[i][k]);
count=max(count,f[i][j][k]);
}
}
}
}
}
return count;
}
};