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At timet, letdp[c1][c2]be the most cherries that we can pick up for two people going from(0, 0)to(r1, c1)and(0, 0)to(r2, c2), wherer1 = t-c1, r2 = t-c2. Our dynamic program proceeds similarly toApproach classSolution {publicintcherryPickup(int[][] grid) {intN =grid.l...
class Solution(object): def cherryPickup(self, grid): """ :type grid: List[List[int]] :rtype: int """ rv = [] self.br_pick(grid, 0, 0, [], rv) print(rv) path = self.pick_max(rv) return len(path) def pick_max(self, paths): m_path = [] for path in paths: if le...
DP(x1, y1, x2, y2)表示第一个人起始在(x1,y1),第二个人起始在(x2,y2)时,能收集到的最多果子 但是这里其实可以压缩一维状态,因为两个人要同时动,而且他们只能向右或者向下动,而且他们最开始的出发位置相同(都为(0,0)点) 因此有 x1+y1 = x2+y2,于是有 y2 = x1+y1-x2,这样就从4维DP变成了...
1463-Cherry-Pickup-II/cpp-1463 CMakeLists.txt main.cpp main2.cpp main3.cpp main4.cpp 1464-Maximum-Product-of-Two-Elements-in-an-Array 1465-Maximum-Area-of-a-Piece-of-Cake-After-Horizontal-and-Vertical-Cuts 1466-Reorder-Routes-to-Make-All-Paths-Lead-to-the-City-Zero 1467-Probability-of...
class Solution{public:intcherryPickup(vector<vector<int>>&grid){intn=grid.size();vector<vector<vector<int>>>dp(n,vector<vector<int>>(n,vector<int>(n,-1)));returnmax(0,dfs(grid,0,0,0,n,dp));}intdfs(vector<vector<int>>&grid,inti,intj,intk,intn,vector<vector<vector<int>>>&...
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public int cherryPickup(int[][] grid) { int ans = 0; int len = grid[0].length; Integer[][][] dp = new Integer[2][len][len]; // null即为不可达 dp[0][0][len - 1] = grid[0][0] + grid[0][len - 1]; for (int row = 1; row < grid.length; row++) { ...
1463. Cherry Pickup II # 题目 # Given a rows x cols matrix grid representing a field of cherries. Each cell in grid represents the number of cherries that you can collect. You have two robots that can collect cherries for you, Robot #1 is located at th
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