LWC 61:741. Cherry Pickup

LWC 61:741. Cherry Pickup

传送门:741. Cherry Pickup

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.

思路: 先出再回来。实际上可以看作两条出发的线抵达N-1即可。此处的trick在于被选择后的状态该如何记录。实际上如果去的时候两条路线重合的话只会加一次cherry。此题采用递归+记忆化的手段,当然因为借鉴了别人的思路,所以具体的解题演变过程在此处就无法展示了。在代码中列举一些自己的comment吧。

Java版本:

    public int cherryPickup(int[][] grid) {
        n = grid.length;
        m = grid[0].length;

        v = new int[n][m];

        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                v[i][j] = grid[i][j];
            }
        }

        mem = new int[52][52][52];
        for (int i = 0; i < 52; ++i) {
            for (int j = 0; j < 52; ++j) {
                for (int k = 0; k < 52; ++k) {
                    mem[i][j][k] = -1;
                }
            }
        }

        // grid[0][0] 始终会被访问到,f在求解问题时,是从起点出发(不包含起点)的最大cherry数
        int ans = f(0, 0, 0, 0) + grid[0][0];
        return ans < -10000 ? 0 : ans;
    }

    int[][] v;
    int n, m;
    int INF = 0x3f3f3f3f;

    int[][][] mem;

    int[] dx = {1, 0};
    int[] dy = {0, 1};
    int f(int x1, int y1, int x2, int y2) {
        // 一旦抵达终态则范围0, 因为此处f的最大cherry数不包含起点,所以终态才为 (n - 1, m - 1)
        if (x1 == n - 1 && y1 == m - 1) return 0;
        // x2一旦确定,y2跟着确定,没必要记录y2的状态
        if (mem[x1][y1][x2] != -1) return mem[x1][y1][x2];
        int ans = -INF;

        for (int i = 0; i < 2; ++i) {
            for (int j = 0; j < 2; ++j) {
                // 只能往右 和 往下
                int nx1 = x1 + dx[i];
                int ny1 = y1 + dy[i];

                int nx2 = x2 + dx[j];
                int ny2 = y2 + dy[j];

                if (nx1 >= 0 && nx1 < n && nx2 >= 0 && nx2 < n && ny1 >= 0 && ny1 < m && ny2 >= 0 && ny2 < m) {
                    if (v[nx1][ny1] == -1) continue;
                    if (v[nx2][ny2] == -1) continue;

                    if (nx1 == nx2 && ny1 == ny2) {
                        // 重合的情况,只需加一次cherry,第一次取cherry,第二次经过但无cherry可取
                        ans = Math.max(ans, v[nx1][ny1] + f(nx1, ny1, nx2, ny2));
                    }
                    else {
                        // 去取一次cherry,回也取一次cherry
                        ans = Math.max(ans, v[nx1][ny1] + v[nx2][ny2] + f(nx1, ny1, nx2, ny2));
                    }
                }
            }
        }
        return mem[x1][y1][x2] = ans;
    }

Python 版本:(超时)

    def cherryPickup(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        n = len(grid)
        m = len(grid[0])

        dp = dict()
        dx = [1, 0]
        dy = [0, 1]

        INF = 0x3f3f3f3f;

        def f(x1, y1, x2, y2):
            if (x1 == n - 1 and y1 == m - 1): return 0
            if ((x1, y1, x2) in dp): return dp[(x1, y1, x2)]

            ans = -INF
            for i in range(2):
                for j in range(2):
                    nx1 = x1 + dx[i]
                    ny1 = y1 + dy[i]

                    nx2 = x2 + dx[j]
                    ny2 = y2 + dy[j]

                    if (nx1 >= 0 and nx1 < n and ny1 >= 0 and ny1 < m and nx2 >= 0 and nx2 < n and ny2 >= 0 and ny2 < m):
                        if (grid[nx1][ny1] == -1): continue
                        if (grid[nx2][ny2] == -1): continue

                        if (nx1 == nx2 and ny1 == ny2):
                            ans = max(ans, grid[nx1][ny1] + f(nx1, ny1, nx2, ny2))
                        else:
                            ans = max(ans, grid[nx1][ny1] + grid[nx2][ny2] + f(nx1, ny1, nx2, ny2))

            dp[(x1, y1, x2)] = ans
            return ans

        ans = f(0, 0, 0, 0) + grid[0][0]
        if ans < -10000: return 0
        return ans        

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