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：top down dp，想最後一步，兩個機器人假設從 1 ~ n - 1全部已經搞定了，那麼從當前往下，會有9種可能，下去。取9種可能性裏面最大的，往下走。如果c1 == c2, 那麼就加一個值，如果c1 != c2, 那麼 就收集兩個，加兩個cell的值；注意dp array是三維的，而且是用Integer來表示，可以dp[row][c1][c2] != null, return;
class Solution {
// Time: 9 * n * m ^2;
// Space: O(n * m ^2);
public int cherryPickup(int[][] grid) {
if(grid == null || grid.length == 0 || grid[0].length == 0) {
return 0;
}
int n = grid.length;
int m = grid[0].length;
Integer[][][] dp = new Integer[n][m][m];
return dfs(grid, n, m, 0, 0, m - 1, dp);
}
private int dfs(int[][] grid, int n, int m, int row, int c1, int c2, Integer[][][] dp) {
if(row == n) {
return 0;
}
if(dp[row][c1][c2] != null) {
return dp[row][c1][c2];
}
int nextMax = 0;
for(int i = -1; i <= 1; i++) {
for(int j = -1; j <= 1; j++) {
int nc1 = c1 + i;
int nc2 = c2 + j;
if(0 <= nc1 && nc1 < m && 0 <= nc2 && nc2 < m) {
nextMax = Math.max(nextMax, dfs(grid, n, m, row + 1, nc1, nc2, dp));
}
}
}
if(c1 == c2) {
dp[row][c1][c2] = nextMax + grid[row][c1];
} else {
// c1 != c2;
dp[row][c1][c2] = nextMax + grid[row][c1] + grid[row][c2];
}
return dp[row][c1][c2];
}
}