Problem
In a N x N grid representing a field of cherries, each cell is one of three possible integers.
0 means the cell is empty, so you can pass through;
1 means the cell contains a cherry, that you can pick up and pass through;
-1 means the cell contains a thorn that blocks your way.
Your task is to collect maximum number of cherries possible by following the rules below:
Starting at the position (0, 0) and reaching (N-1, N-1) by moving right or down through valid path cells (cells with value 0 or 1);
After reaching (N-1, N-1), returning to (0, 0) by moving left or up through valid path cells;
When passing through a path cell containing a cherry, you pick it up and the cell becomes an empty cell (0);
If there is no valid path between (0, 0) and (N-1, N-1), then no cherries can be collected.
Example 1:
Input: grid =
[[0, 1, -1],
[1, 0, -1],
[1, 1, 1]]
Output: 5
Explanation:
The player started at (0, 0) and went down, down, right right to reach (2, 2).
4 cherries were picked up during this single trip, and the matrix becomes [[0,1,-1],[0,0,-1],[0,0,0]].
Then, the player went left, up, up, left to return home, picking up one more cherry.
The total number of cherries picked up is 5, and this is the maximum possible.Note:
grid is an N by N 2D array, with 1 <= N <= 50.
Each grid[i][j] is an integer in the set {-1, 0, 1}.
It is guaranteed that grid[0][0] and grid[N-1][N-1] are not -1.
Solution
two men goes at the same time
x1+y1 = x2+y2
class Solution {
public int cherryPickup(int[][] grid) {
int n = grid.length;
int[][][] dp = new int[n+1][n+1][n+1];
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= n; j++) {
Arrays.fill(dp[i][j], Integer.MIN_VALUE);
}
}
dp[1][1][1] = grid[0][0];
for (int x1 = 1; x1 <= n; x1++) {
for (int y1 = 1; y1 <= n; y1++) {
for (int x2 = 1; x2 <= n; x2++) {
int y2 = x1+y1-x2;
if (dp[x1][y1][x2] > 0 ||
y2 < 1 || y2 > n ||
grid[x1-1][y1-1] == -1 || grid[x2-1][y2-1] == -1) continue;
int preMax = Math.max(
Math.max(dp[x1-1][y1][x2], dp[x1-1][y1][x2-1]),
Math.max(dp[x1][y1-1][x2], dp[x1][y1-1][x2-1])
);
if (preMax < 0) continue;
dp[x1][y1][x2] = preMax + grid[x1-1][y1-1];
if (x1 != x2) dp[x1][y1][x2] += grid[x2-1][y2-1];
}
}
}
return dp[n][n][n] < 0 ? 0 : dp[n][n][n];
}
}