[Leetcode] 741. Cherry Pickup 解题报告

题目：

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：

  我们定义dp[i][j]表示从(0, 0)开始的两条长为k的路径上，可以摘到的草莓的最大个数。而这两条长为k的路径的最终点为(i, k - i)和(j, k - j)。然后我们分别计算两条路径上可以摘到的草莓数。后面的没怎么看懂了。。。等到完全理解了思路再补充吧。。。

  代码：

  class Solution {
  public:
      int cherryPickup(vector<vector<int>>& grid) {
          int n = grid.size();
          /* dp holds maximum # of cherries two k-length paths can pickup. 
          The two k-length paths arrive at (i, k - i) and (j, k - j), respectively. */
          vector<vector<int>> dp(n, vector<int>(n, -1));
          dp[0][0] = grid[0][0];              // length k = 0
          const int maxK = 2 * (n - 1);       // maxK: number of steps from (0, 0) to (n-1, n-1).
          for (int k = 1; k <= maxK; ++k) {   // for every length k
              vector<vector<int>> curr(n, vector<int>(n, -1));
              // one path of length k arrive at (i, k - i)
              for (int i = 0; i < n && i <= k; ++i) {
                  if ( k - i >= n) {
                      continue;
                  }
                  // another path of length k arrive at (j, k - j)
                  for (int j = 0; j < n && j <= k; ++j) {
                      if (k - j >= n) {
                          continue;
                      }
                      if (grid[i][k - i] < 0 || grid[j][k - j] < 0) {     // keep away from thorns
                          continue;
                      }
                      int cherries = dp[i][j];    // # of cherries picked up by the two (k-1)-length paths.
                      if (i > 0) {
                          cherries = std::max(cherries, dp[i - 1][j]);
                      }
                      if (j > 0) {
                          cherries = std::max(cherries, dp[i][j - 1]);
                      }
                      if (i > 0 && j > 0) {
                          cherries = std::max(cherries, dp[i - 1][j - 1]);
                      }
                      // No viable way to arrive at (i, k - i)-(j, k-j).
                      if (cherries < 0) {
                          continue;
                      }
                      // Pickup cherries at (i, k - i) and (j, k -j ) if i != j. Otherwise, pickup (i, k-i).
                      cherries += grid[i][k - i] + (i == j ? 0 : grid[j][k - j]);
                      curr[i][j] = cherries;
                  }
              }
              dp = std::move(curr); 
          }
          return std::max(dp[n-1][n-1], 0); 
      }
  };