LeetCode741. Cherry Pickup

https://leetcode.com/problems/cherry-pickup/description/

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.

思路

解答给出的第一种方法时贪心，但是并不是正确答案，正确解法还是万能的dp。

在 t 个steps后，我们到的位置 (r, c)，有 r+c=t 。如果有两个人在位置在 t 个steps后在位置 positions (r1, c1) and (r2, c2) 上，那么有 r2 = r1 + c1 - c2。 这意味着变量r1, c1, c2唯一决定了两个都走了r1 + c1 number of steps人的位置。这是我们动态规划思想的基础。

自顶向下的dp：

Let dp[r1][c1][c2] be the most number of cherries obtained by two people starting at (r1, c1) and (r2, c2)and walking towards (N-1, N-1) picking up cherries, where r2 = r1+c1-c2.

If grid[r1][c1] and grid[r2][c2] are not thorns, then the value of dp[r1][c1][c2] is (grid[r1][c1] + grid[r2][c2]), plus the maximum of dp[r1+1][c1][c2], dp[r1][c1+1][c2], dp[r1+1][c1][c2+1], dp[r1][c1+1][c2+1] as appropriate. We should also be careful to not double count in case (r1, c1) == (r2, c2).

Why did we say it was the maximum of dp[r+1][c1][c2] etc.? It corresponds to the 4 possibilities for person 1 and 2 moving down and right:

• Person 1 down and person 2 down: dp[r1+1][c1][c2];
• Person 1 right and person 2 down: dp[r1][c1+1][c2];
• Person 1 down and person 2 right: dp[r1+1][c1][c2+1];
• Person 1 right and person 2 right: dp[r1][c1+1][c2+1];

要点：1. 将题目中要求的从起始到末尾在返回起始点，等价为二个人同时从起点出发去重点。

　　    2. 一个三维的dp数组，标记了两个人的位置，以及当前最优解

　　　3. 子问题之间的关系，要避免重复计算。

代码

class Solution {
    int[][][] memo;
    int[][] grid;
    int N;
    public int cherryPickup(int[][] grid) {
        this.grid = grid;
        N = grid.length;
        memo = new int[N][N][N];
        for (int[][] layer: memo)
            for (int[] row: layer)
                Arrays.fill(row, Integer.MIN_VALUE);
        return Math.max(0, dp(0, 0, 0));
    }
    public int dp(int r1, int c1, int c2) {
        int r2 = r1 + c1 - c2;
        if (N == r1 || N == r2 || N == c1 || N == c2 ||
                grid[r1][c1] == -1 || grid[r2][c2] == -1) {　//到达边界或者遇到阻碍
            return -999999;
        } else if (r1 == N-1 && c1 == N-1) {
            return grid[r1][c1];
        } else if (memo[r1][c1][c2] != Integer.MIN_VALUE) {　// 如果这个位置计算过了则不需要再次计算
            return memo[r1][c1][c2];
        } else {
            int ans = grid[r1][c1];
            if (c1 != c2) ans += grid[r2][c2];
            ans += Math.max(Math.max(dp(r1, c1+1, c2+1), dp(r1+1, c1, c2+1)),
                            Math.max(dp(r1, c1+1, c2), dp(r1+1, c1, c2)));
            memo[r1][c1][c2] = ans;
            return ans;
        }
    }
}

上面是自顶向下的dp。另一中是自低向上的dp:

At time t, let dp[c1][c2] be the most cherries that we can pick up for two people going from (0, 0) to (r1, c1)and (0, 0) to (r2, c2), where r1 = t-c1, r2 = t-c2. Our dynamic program proceeds similarly to Approach 

class Solution {
    public int cherryPickup(int[][] grid) {
        int N = grid.length;
        int[][] dp = new int[N][N];
        for (int[] row: dp) Arrays.fill(row, Integer.MIN_VALUE);
        dp[0][0] = grid[0][0];

        for (int t = 1; t <= 2*N - 2; ++t) {
            int[][] dp2 = new int[N][N];
            for (int[] row: dp2) Arrays.fill(row, Integer.MIN_VALUE);

            for (int i = Math.max(0, t-(N-1)); i <= Math.min(N-1, t); ++i) {
                for (int j = Math.max(0, t-(N-1)); j <= Math.min(N-1, t); ++j) {
                    if (grid[i][t-i] == -1 || grid[j][t-j] == -1) continue;
                    int val = grid[i][t-i];
                    if (i != j) val += grid[j][t-j];
                    for (int pi = i-1; pi <= i; ++pi)
                        for (int pj = j-1; pj <= j; ++pj)
                            if (pi >= 0 && pj >= 0)
                                dp2[i][j] = Math.max(dp2[i][j], dp[pi][pj] + val);
                }
            }
            dp = dp2;
        }
        return Math.max(0, dp[N-1][N-1]);
    }
}