[LeetCode] 741. Cherry Pickup 捡樱桃

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.

和64. Minimum Path Sum 类似，但这个题还要返回到起点，而且捡过的位置由1变为0，如果分别计算去时和回来时的路径，就要把捡过的变为0，会很麻烦。

解法：DP, 同时计算去和回的dp值。

参考：Grandyang

https://leetcode.com/problems/cherry-pickup/discuss/109903/Step-by-step-guidance-of-the-O(N3)-time-and-O(N2)-space-solution

Java:

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public int cherryPickup(int[][] grid) {     int N = grid.length, M = (N << 1) - 1;     int[][] dp = new int[N][N];     dp[0][0] = grid[0][0];               for (int n = 1; n < M; n++) {     for (int i = N - 1; i >= 0; i--) {         for (int p = N - 1; p >= 0; p--) {         int j = n - i, q = n - p;                           if (j < 0 || j >= N || q < 0 || q >= N || grid[i][j] < 0 || grid[p][q] < 0) {                     dp[i][p] = -1;                     continue;                  }                     if (i > 0) dp[i][p] = Math.max(dp[i][p], dp[i - 1][p]);          if (p > 0) dp[i][p] = Math.max(dp[i][p], dp[i][p - 1]);          if (i > 0 && p > 0) dp[i][p] = Math.max(dp[i][p], dp[i - 1][p - 1]);                     if (dp[i][p] >= 0) dp[i][p] += grid[i][j] + (i != p ? grid[p][q] : 0)              }      }     }           return Math.max(dp[N - 1][N - 1], 0); }

Python:

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class Solution(object):     def cherryPickup(self, grid):         """         :type grid: List[List[int]]         :rtype: int         """         # dp holds the max # of cherries two k-length paths can pickup.         # The two k-length paths arrive at (i, k - i) and (j, k - j),         # respectively.         n = len(grid)         dp = [[-1 for _ in xrange(n)] for _ in xrange(n)]         dp[0][0] = grid[0][0]         max_len = 2 * (n-1)         directions = [(0, 0), (-1, 0), (0, -1), (-1, -1)]         for k in xrange(1, max_len+1):             for i in reversed(xrange(max(0, k-n+1), min(k+1, n))):  # 0 <= i < n, 0 <= k-i < n                 for j in reversed(xrange(i, min(k+1, n))):          # i <= j < n, 0 <= k-j < n                     if grid[i][k-i] == -1 or grid[j][k-j] == -1:                         dp[i][j] = -1                         continue                     cnt = grid[i][k-i]                     if i != j:                         cnt += grid[j][k-j]                     max_cnt = -1                     for direction in directions:                         ii, jj = i+direction[0], j+direction[1]                         if ii >= 0 and jj >= 0 and dp[ii][jj] >= 0:                             max_cnt = max(max_cnt, dp[ii][jj]+cnt)                     dp[i][j] = max_cnt         return max(dp[n-1][n-1], 0)

Python:

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class Solution(object):     def cherryPickup(self, grid):         """         :type grid: List[List[int]]         :rtype: int         """         if grid[-1][-1] == -1: return 0                   # set up cache         self.grid = grid         self.memo = {}         self.N = len(grid)                   return max(self.dp(0, 0, 0, 0), 0)               def dp(self, i1, j1, i2, j2):         # already stored: return         if (i1, j1, i2, j2) in self.memo: return self.memo[(i1, j1, i2, j2)]                   # end states: 1. out of grid 2. at the right bottom corner 3. hit a thorn         N = self.N         if i1 == N or j1 == N or i2 == N or j2 == N: return -1         if i1 == N-1 and j1 == N-1 and i2 == N-1 and j2 == N-1: return self.grid[-1][-1]         if self.grid[i1][j1] == -1 or self.grid[i2][j2] == -1: return -1                   # now can take a step in two directions at each end, which amounts to 4 combinations in total         dd = self.dp(i1+1, j1, i2+1, j2)         dr = self.dp(i1+1, j1, i2, j2+1)         rd = self.dp(i1, j1+1, i2+1, j2)         rr = self.dp(i1, j1+1, i2, j2+1)         maxComb = max([dd, dr, rd, rr])                   # find if there is a way to reach the end         if maxComb == -1:             out = -1         else:             # same cell, can only count this cell once             if i1 == i2 and j1 == j2:                 out = maxComb + self.grid[i1][j1]             # different cell, can collect both             else:                 out = maxComb + self.grid[i1][j1] + self.grid[i2][j2]                           # cache result         self.memo[(i1, j1, i2, j2)] = out         return out

C++:

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class Solution { public:     int cherryPickup(vector<vector<int>>& grid) {         int n = grid.size(), mx = 2 * n - 1;         vector<vector<int>> dp(n, vector<int>(n, -1));         dp[0][0] = grid[0][0];         for (int k = 1; k < mx; ++k) {             for (int i = n - 1; i >= 0; --i) {                 for (int p = n - 1; p >= 0; --p) {                     int j = k - i, q = k - p;                     if (j < 0 || j >= n || q < 0 || q >= n || grid[i][j] < 0 || grid[p][q] < 0) {                         dp[i][p] = -1;                         continue;                     }                     if (i > 0) dp[i][p] = max(dp[i][p], dp[i - 1][p]);                     if (p > 0) dp[i][p] = max(dp[i][p], dp[i][p - 1]);                     if (i > 0 && p > 0) dp[i][p] = max(dp[i][p], dp[i - 1][p - 1]);                     if (dp[i][p] >= 0) dp[i][p] += grid[i][j] + (i != p ? grid[p][q] : 0);                 }             }         }         return max(dp[n - 1][n - 1], 0);     } };

　　

 

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