741. Cherry Pickup

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

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.

N阶方阵上0代表可通过，1代表有果子，-1代表不可通过，从(0, 0)开始可每次向下或向右走到(N - 1, N-1)，再每步向上或向左走回到(0，0)，问最多能拿到多少果子。往回走可以看作是另一条正着走的path。但用两遍max path sum的谈心法是错误的，因为第二遍会取决于第一遍时的结果，而第一遍时最优解不一定是全局最优。既然问题出在了第二遍时丢失了第一遍时的信息，那把两条paths看作是两个人同时独立走，并在它俩相遇时只拿一次果子：把两条paths的位置信息encode进DP状态，就避免了第二条path错误的只关联在第一条path的最优解上。action分别为第一个人下/右与第二个人下右四种，dp[x1][y1][x2]表示从(x1, y1, x2, x1 + y1 - x2)出发到(N-1, N-1)能拿到最多果子数。

思路参考自这。

/*
 * @lc app=leetcode id=741 lang=cpp
 *
 * [741] Cherry Pickup
 */

// @lc code=start
class Solution {
public:
    int N;
    int dfs(vector<vector<int>> &g, vector<vector<vector<int>>> &dp, int x1, int y1, int x2) {
        const int y2 = x1 + y1 - x2;
        if (x1 == N || y1 == N || x2 == N || y2 == N || g[x1][y1] == -1 || g[x2][y2] == -1) return -1; 
        if (dp[x1][y1][x2] != INT_MIN) return dp[x1][y1][x2];
        int res = g[x1][y1];
        if (x1 != x2) res += g[x2][y2];
        if (x1 == N - 1 && y1 == N - 1) return res;
        const auto r = max(dfs(g, dp, x1 + 1, y1, x2 + 1), max(dfs(g, dp, x1, y1 + 1, x2 + 1), max(dfs(g, dp, x1 + 1, y1, x2), dfs(g, dp, x1, y1 + 1, x2))));
        if (r == -1) res = -1;
        else res += r;
        return dp[x1][y1][x2] = res;
    }

    int cherryPickup(vector<vector<int>>& grid) {
        N = grid.size();
        auto dp = vector<vector<vector<int>>>(N, vector<vector<int>>(N, vector<int>(N, INT_MIN)));
        return max(0, dfs(grid, dp, 0, 0, 0));
    }
};
// @lc code=end