Question

1463. 摘樱桃 II

English Version

题目描述

给你一个 rows x cols 的矩阵 grid 来表示一块樱桃地。 grid 中每个格子的数字表示你能获得的樱桃数目。

你有两个机器人帮你收集樱桃，机器人 1 从左上角格子 (0,0) 出发，机器人 2 从右上角格子 (0, cols-1) 出发。

请你按照如下规则，返回两个机器人能收集的最多樱桃数目：

• 从格子 (i,j) 出发，机器人可以移动到格子 (i+1, j-1)，(i+1, j) 或者 (i+1, j+1) 。
• 当一个机器人经过某个格子时，它会把该格子内所有的樱桃都摘走，然后这个位置会变成空格子，即没有樱桃的格子。
• 当两个机器人同时到达同一个格子时，它们中只有一个可以摘到樱桃。
• 两个机器人在任意时刻都不能移动到 grid 外面。
• 两个机器人最后都要到达 grid 最底下一行。

 

示例 1：

输入：grid = [[3,1,1],[2,5,1],[1,5,5],[2,1,1]]
输出：24
解释：机器人 1 和机器人 2 的路径在上图中分别用绿色和蓝色表示。
机器人 1 摘的樱桃数目为 (3 + 2 + 5 + 2) = 12 。
机器人 2 摘的樱桃数目为 (1 + 5 + 5 + 1) = 12 。
樱桃总数为： 12 + 12 = 24 。

示例 2：

输入：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]]
输出：28
解释：机器人 1 和机器人 2 的路径在上图中分别用绿色和蓝色表示。
机器人 1 摘的樱桃数目为 (1 + 9 + 5 + 2) = 17 。
机器人 2 摘的樱桃数目为 (1 + 3 + 4 + 3) = 11 。
樱桃总数为： 17 + 11 = 28 。

示例 3：

输入：grid = [[1,0,0,3],[0,0,0,3],[0,0,3,3],[9,0,3,3]]
输出：22

示例 4：

输入：grid = [[1,1],[1,1]]
输出：4

 

提示：

• rows == grid.length
• cols == grid[i].length
• 2 <= rows, cols <= 70
• 0 <= grid[i][j] <= 100 

解法

方法一：动态规划

我们定义 $f[i][j_1][j_2]$ 表示两个机器人分别在第 $i$ 行的位置 $j_1$ 和 $j_2$ 时能够摘到的最多樱桃数目。初始时 $f[0][0][n-1] = grid[0][0] + grid[0][n-1]$，其余值均为 $-1$。答案为 $\max_{0 \leq j_1, j_2 < n} f[m-1][j_1][j_2]$。

考虑 $f[i][j_1][j_2]$，如果 $j_1 \neq j_2$，那么机器人在第 $i$ 行能摘到的樱桃数目为 $grid[i][j_1] + grid[i][j_2]$；如果 $j_1 = j_2$，那么机器人在第 $i$ 行能摘到的樱桃数目为 $grid[i][j_1]$。我们可以枚举两个机器人的上一个状态 $f[i-1][y1][y2]$，其中 $y_1, y_2$ 分别是两个机器人在第 $i-1$ 行的位置，那么有 $y_1 \in {j_1-1, j_1, j_1+1}$ 且 $y_2 \in {j_2-1, j_2, j_2+1}$。状态转移方程如下：

$$ f[i][j_1][j_2] = \max_{y_1 \in {j_1-1, j_1, j_1+1}, y_2 \in {j_2-1, j_2, j_2+1}} f[i-1][y_1][y_2] + \begin{cases} grid[i][j_1] + grid[i][j_2], & j_1 \neq j_2 \ grid[i][j_1], & j_1 = j_2 \end{cases} $$

其中 $f[i-1][y_1][y_2]$ 为 $-1$ 时需要忽略。

最终的答案即为 $\max_{0 \leq j_1, j_2 < n} f[m-1][j_1][j_2]$。

时间复杂度 $O(m \times n^2)$，空间复杂度 $O(m \times n^2)$。其中 $m$ 和 $n$ 分别是网格的行数和列数。

注意到 $f[i][j_1][j_2]$ 的计算只和 $f[i-1][y_1][y_2]$ 有关，因此我们可以使用滚动数组优化空间复杂度，空间复杂度优化后的时间复杂度为 $O(n^2)$。

class Solution:
  def cherryPickup(self, grid: List[List[int]]) -> int:
    m, n = len(grid), len(grid[0])
    f = [[[-1] * n for _ in range(n)] for _ in range(m)]
    f[0][0][n - 1] = grid[0][0] + grid[0][n - 1]
    for i in range(1, m):
      for j1 in range(n):
        for j2 in range(n):
          x = grid[i][j1] + (0 if j1 == j2 else grid[i][j2])
          for y1 in range(j1 - 1, j1 + 2):
            for y2 in range(j2 - 1, j2 + 2):
              if 0 <= y1 < n and 0 <= y2 < n and f[i - 1][y1][y2] != -1:
                f[i][j1][j2] = max(f[i][j1][j2], f[i - 1][y1][y2] + x)
    return max(f[-1][j1][j2] for j1, j2 in product(range(n), range(n)))

class Solution {
  public int cherryPickup(int[][] grid) {
    int m = grid.length, n = grid[0].length;
    int[][][] f = new int[m][n][n];
    for (var g : f) {
      for (var h : g) {
        Arrays.fill(h, -1);
      }
    }
    f[0][0][n - 1] = grid[0][0] + grid[0][n - 1];
    for (int i = 1; i < m; ++i) {
      for (int j1 = 0; j1 < n; ++j1) {
        for (int j2 = 0; j2 < n; ++j2) {
          int x = grid[i][j1] + (j1 == j2 ? 0 : grid[i][j2]);
          for (int y1 = j1 - 1; y1 <= j1 + 1; ++y1) {
            for (int y2 = j2 - 1; y2 <= j2 + 1; ++y2) {
              if (y1 >= 0 && y1 < n && y2 >= 0 && y2 < n && f[i - 1][y1][y2] != -1) {
                f[i][j1][j2] = Math.max(f[i][j1][j2], f[i - 1][y1][y2] + x);
              }
            }
          }
        }
      }
    }
    int ans = 0;
    for (int j1 = 0; j1 < n; ++j1) {
      for (int j2 = 0; j2 < n; ++j2) {
        ans = Math.max(ans, f[m - 1][j1][j2]);
      }
    }
    return ans;
  }
}

class Solution {
public:
  int cherryPickup(vector<vector<int>>& grid) {
    int m = grid.size(), n = grid[0].size();
    int f[m][n][n];
    memset(f, -1, sizeof(f));
    f[0][0][n - 1] = grid[0][0] + grid[0][n - 1];
    for (int i = 1; i < m; ++i) {
      for (int j1 = 0; j1 < n; ++j1) {
        for (int j2 = 0; j2 < n; ++j2) {
          int x = grid[i][j1] + (j1 == j2 ? 0 : grid[i][j2]);
          for (int y1 = j1 - 1; y1 <= j1 + 1; ++y1) {
            for (int y2 = j2 - 1; y2 <= j2 + 1; ++y2) {
              if (y1 >= 0 && y1 < n && y2 >= 0 && y2 < n && f[i - 1][y1][y2] != -1) {
                f[i][j1][j2] = max(f[i][j1][j2], f[i - 1][y1][y2] + x);
              }
            }
          }
        }
      }
    }
    int ans = 0;
    for (int j1 = 0; j1 < n; ++j1) {
      for (int j2 = 0; j2 < n; ++j2) {
        ans = max(ans, f[m - 1][j1][j2]);
      }
    }
    return ans;
  }
};

func cherryPickup(grid [][]int) int {
  m, n := len(grid), len(grid[0])
  f := make([][][]int, m)
  for i := range f {
    f[i] = make([][]int, n)
    for j := range f[i] {
      f[i][j] = make([]int, n)
      for k := range f[i][j] {
        f[i][j][k] = -1
      }
    }
  }
  f[0][0][n-1] = grid[0][0] + grid[0][n-1]
  for i := 1; i < m; i++ {
    for j1 := 0; j1 < n; j1++ {
      for j2 := 0; j2 < n; j2++ {
        x := grid[i][j1]
        if j1 != j2 {
          x += grid[i][j2]
        }
        for y1 := j1 - 1; y1 <= j1+1; y1++ {
          for y2 := j2 - 1; y2 <= j2+1; y2++ {
            if y1 >= 0 && y1 < n && y2 >= 0 && y2 < n && f[i-1][y1][y2] != -1 {
              f[i][j1][j2] = max(f[i][j1][j2], f[i-1][y1][y2]+x)
            }
          }
        }
      }
    }
  }
  ans := 0
  for j1 := 0; j1 < n; j1++ {
    for j2 := 0; j2 < n; j2++ {
      ans = max(ans, f[m-1][j1][j2])
    }
  }
  return ans
}

function cherryPickup(grid: number[][]): number {
  const m = grid.length;
  const n = grid[0].length;
  const f: number[][][] = new Array(m)
    .fill(0)
    .map(() => new Array(n).fill(0).map(() => new Array(n).fill(-1)));
  f[0][0][n - 1] = grid[0][0] + grid[0][n - 1];
  for (let i = 1; i < m; ++i) {
    for (let j1 = 0; j1 < n; ++j1) {
      for (let j2 = 0; j2 < n; ++j2) {
        const x = grid[i][j1] + (j1 === j2 ? 0 : grid[i][j2]);
        for (let y1 = j1 - 1; y1 <= j1 + 1; ++y1) {
          for (let y2 = j2 - 1; y2 <= j2 + 1; ++y2) {
            if (y1 >= 0 && y1 < n && y2 >= 0 && y2 < n && f[i - 1][y1][y2] !== -1) {
              f[i][j1][j2] = Math.max(f[i][j1][j2], f[i - 1][y1][y2] + x);
            }
          }
        }
      }
    }
  }
  let ans = 0;
  for (let j1 = 0; j1 < n; ++j1) {
    for (let j2 = 0; j2 < n; ++j2) {
      ans = Math.max(ans, f[m - 1][j1][j2]);
    }
  }
  return ans;
}

方法二

class Solution:
  def cherryPickup(self, grid: List[List[int]]) -> int:
    m, n = len(grid), len(grid[0])
    f = [[-1] * n for _ in range(n)]
    g = [[-1] * n for _ in range(n)]
    f[0][n - 1] = grid[0][0] + grid[0][n - 1]
    for i in range(1, m):
      for j1 in range(n):
        for j2 in range(n):
          x = grid[i][j1] + (0 if j1 == j2 else grid[i][j2])
          for y1 in range(j1 - 1, j1 + 2):
            for y2 in range(j2 - 1, j2 + 2):
              if 0 <= y1 < n and 0 <= y2 < n and f[y1][y2] != -1:
                g[j1][j2] = max(g[j1][j2], f[y1][y2] + x)
      f, g = g, f
    return max(f[j1][j2] for j1, j2 in product(range(n), range(n)))

class Solution {
  public int cherryPickup(int[][] grid) {
    int m = grid.length, n = grid[0].length;
    int[][] f = new int[n][n];
    int[][] g = new int[n][n];
    for (int i = 0; i < n; ++i) {
      Arrays.fill(f[i], -1);
      Arrays.fill(g[i], -1);
    }
    f[0][n - 1] = grid[0][0] + grid[0][n - 1];
    for (int i = 1; i < m; ++i) {
      for (int j1 = 0; j1 < n; ++j1) {
        for (int j2 = 0; j2 < n; ++j2) {
          int x = grid[i][j1] + (j1 == j2 ? 0 : grid[i][j2]);
          for (int y1 = j1 - 1; y1 <= j1 + 1; ++y1) {
            for (int y2 = j2 - 1; y2 <= j2 + 1; ++y2) {
              if (y1 >= 0 && y1 < n && y2 >= 0 && y2 < n && f[y1][y2] != -1) {
                g[j1][j2] = Math.max(g[j1][j2], f[y1][y2] + x);
              }
            }
          }
        }
      }
      int[][] t = f;
      f = g;
      g = t;
    }
    int ans = 0;
    for (int j1 = 0; j1 < n; ++j1) {
      for (int j2 = 0; j2 < n; ++j2) {
        ans = Math.max(ans, f[j1][j2]);
      }
    }
    return ans;
  }
}

class Solution {
public:
  int cherryPickup(vector<vector<int>>& grid) {
    int m = grid.size(), n = grid[0].size();
    vector<vector<int>> f(n, vector<int>(n, -1));
    vector<vector<int>> g(n, vector<int>(n, -1));
    f[0][n - 1] = grid[0][0] + grid[0][n - 1];
    for (int i = 1; i < m; ++i) {
      for (int j1 = 0; j1 < n; ++j1) {
        for (int j2 = 0; j2 < n; ++j2) {
          int x = grid[i][j1] + (j1 == j2 ? 0 : grid[i][j2]);
          for (int y1 = j1 - 1; y1 <= j1 + 1; ++y1) {
            for (int y2 = j2 - 1; y2 <= j2 + 1; ++y2) {
              if (y1 >= 0 && y1 < n && y2 >= 0 && y2 < n && f[y1][y2] != -1) {
                g[j1][j2] = max(g[j1][j2], f[y1][y2] + x);
              }
            }
          }
        }
      }
      swap(f, g);
    }
    int ans = 0;
    for (int j1 = 0; j1 < n; ++j1) {
      for (int j2 = 0; j2 < n; ++j2) {
        ans = max(ans, f[j1][j2]);
      }
    }
    return ans;
  }
};

func cherryPickup(grid [][]int) int {
  m, n := len(grid), len(grid[0])
  f := make([][]int, n)
  g := make([][]int, n)
  for i := range f {
    f[i] = make([]int, n)
    g[i] = make([]int, n)
    for j := range f[i] {
      f[i][j] = -1
      g[i][j] = -1
    }
  }
  f[0][n-1] = grid[0][0] + grid[0][n-1]
  for i := 1; i < m; i++ {
    for j1 := 0; j1 < n; j1++ {
      for j2 := 0; j2 < n; j2++ {
        x := grid[i][j1]
        if j1 != j2 {
          x += grid[i][j2]
        }
        for y1 := j1 - 1; y1 <= j1+1; y1++ {
          for y2 := j2 - 1; y2 <= j2+1; y2++ {
            if y1 >= 0 && y1 < n && y2 >= 0 && y2 < n && f[y1][y2] != -1 {
              g[j1][j2] = max(g[j1][j2], f[y1][y2]+x)
            }
          }
        }
      }
    }
    f, g = g, f
  }
  ans := 0
  for j1 := 0; j1 < n; j1++ {
    for j2 := 0; j2 < n; j2++ {
      ans = max(ans, f[j1][j2])
    }
  }
  return ans
}

function cherryPickup(grid: number[][]): number {
  const m = grid.length;
  const n = grid[0].length;
  let f: number[][] = new Array(n).fill(0).map(() => new Array(n).fill(-1));
  let g: number[][] = new Array(n).fill(0).map(() => new Array(n).fill(-1));
  f[0][n - 1] = grid[0][0] + grid[0][n - 1];
  for (let i = 1; i < m; ++i) {
    for (let j1 = 0; j1 < n; ++j1) {
      for (let j2 = 0; j2 < n; ++j2) {
        const x = grid[i][j1] + (j1 === j2 ? 0 : grid[i][j2]);
        for (let y1 = j1 - 1; y1 <= j1 + 1; ++y1) {
          for (let y2 = j2 - 1; y2 <= j2 + 1; ++y2) {
            if (y1 >= 0 && y1 < n && y2 >= 0 && y2 < n && f[y1][y2] !== -1) {
              g[j1][j2] = Math.max(g[j1][j2], f[y1][y2] + x);
            }
          }
        }
      }
    }
    [f, g] = [g, f];
  }
  let ans = 0;
  for (let j1 = 0; j1 < n; ++j1) {
    for (let j2 = 0; j2 < n; ++j2) {
      ans = Math.max(ans, f[j1][j2]);
    }
  }
  return ans;
}