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| 1 | +// Problem(3-D DP) : We are given an ‘N *M’ matrix.Every cell of the matrix has some chocolates on it, mat[i][j] gives us the |
| 2 | +// number of chocolates.We have two friends ‘Alice’ and ‘Bob’.initially, Alice is standing on the cell(0, 0) and Bob is standing |
| 3 | +// on the cell(0, M - 1).Both of them can move only to the cells below them in these three directions : to the bottom cell(↓), |
| 4 | +// to the bottom - right cell(↘), or to the bottom - left cell(↙). When Alica and Bob visit a cell, they take all the chocolates |
| 5 | +// from that cell with them. It can happen that they visit the same cell, in that case, the chocolates need to be considered only |
| 6 | +// once. They cannot go out of the boundary of the given matrix, we need to return the maximum number of chocolates that Bob and |
| 7 | +// Alice can together collect. |
| 8 | + |
| 9 | +#include <bits/stdc++.h> |
| 10 | +using namespace std; |
| 11 | + |
| 12 | +int f(int i, int j1, int j2, vector<vector<int>> &grid, int n, int m, vector<vector<vector<int>>> &dp) |
| 13 | +{ |
| 14 | + if (i == n) |
| 15 | + { |
| 16 | + return 0; |
| 17 | + } |
| 18 | + if (dp[i][j1][j2] != -1) |
| 19 | + { |
| 20 | + return dp[i][j1][j2]; |
| 21 | + } |
| 22 | + int mx = INT_MIN; |
| 23 | + // if Alice can move to the left column in the row below |
| 24 | + if (((j1 - 1) >= 0)) |
| 25 | + { |
| 26 | + // if Bob can move to the same column in the row below |
| 27 | + mx = max(mx, f(i + 1, j1 - 1, j2, grid, n, m, dp)); |
| 28 | + // if Bob can move to the left column in the row below |
| 29 | + if ((j2 - 1) >= 0) |
| 30 | + { |
| 31 | + mx = max(mx, f(i + 1, j1 - 1, j2 - 1, grid, n, m, dp)); |
| 32 | + } |
| 33 | + // if Bob can move to the right column in the row below |
| 34 | + if ((j2 + 1) < m) |
| 35 | + { |
| 36 | + mx = max(mx, f(i + 1, j1 - 1, j2 + 1, grid, n, m, dp)); |
| 37 | + } |
| 38 | + } |
| 39 | + // if Alice can move to the same column in the row below |
| 40 | + // if Bob can move to the same column in the row below |
| 41 | + mx = max(mx, f(i + 1, j1, j2, grid, n, m, dp)); |
| 42 | + // if Bob can move to the left column in the row below |
| 43 | + if ((j2 - 1) >= 0) |
| 44 | + { |
| 45 | + mx = max(mx, f(i + 1, j1, j2 - 1, grid, n, m, dp)); |
| 46 | + } |
| 47 | + // if Bob can move to the right column in the row below |
| 48 | + if ((j2 + 1) < m) |
| 49 | + { |
| 50 | + mx = max(mx, f(i + 1, j1, j2 + 1, grid, n, m, dp)); |
| 51 | + } |
| 52 | + // if Alice can move to the right column in the row below |
| 53 | + if (((j1 + 1) < m)) |
| 54 | + { |
| 55 | + // if Bob can move to the same column in the row below |
| 56 | + mx = max(mx, f(i + 1, j1 + 1, j2, grid, n, m, dp)); |
| 57 | + // if Bob can move to the left column in the row below |
| 58 | + if ((j2 - 1) >= 0) |
| 59 | + { |
| 60 | + mx = max(mx, f(i + 1, j1 + 1, j2 - 1, grid, n, m, dp)); |
| 61 | + } |
| 62 | + // if Bob can move to the right column in the row below |
| 63 | + if ((j2 + 1) < m) |
| 64 | + { |
| 65 | + mx = max(mx, f(i + 1, j1 + 1, j2 + 1, grid, n, m, dp)); |
| 66 | + } |
| 67 | + } |
| 68 | + if (j1 == j2) |
| 69 | + { |
| 70 | + // if same cell for both alice and bob, count chocolates only once |
| 71 | + return dp[i][j1][j2] = (mx + grid[i][j1]); |
| 72 | + } |
| 73 | + else |
| 74 | + { |
| 75 | + return dp[i][j1][j2] = (mx + grid[i][j1] + grid[i][j2]); |
| 76 | + } |
| 77 | +} |
| 78 | + |
| 79 | +int maximumChocolates(int r, int c, vector<vector<int>> &grid) |
| 80 | +{ |
| 81 | + vector<vector<vector<int>>> dp(r, vector<vector<int>>(c, vector<int>(c))); |
| 82 | + for (int i = 0; i < r; i++) |
| 83 | + { |
| 84 | + for (int j = 0; j < c; j++) |
| 85 | + { |
| 86 | + for (int k = 0; k < c; k++) |
| 87 | + { |
| 88 | + dp[i][j][k] = -1; |
| 89 | + } |
| 90 | + } |
| 91 | + } |
| 92 | + return f(0, 0, c - 1, grid, r, c, dp); |
| 93 | +} |
| 94 | + |
| 95 | +int main() |
| 96 | +{ |
| 97 | + int r, c; |
| 98 | + cin >> r >> c; |
| 99 | + vector<vector<int>> grid(r, vector<int>(c)); |
| 100 | + for (int i = 0; i < r; i++) |
| 101 | + { |
| 102 | + for (int j = 0; j < c; j++) |
| 103 | + { |
| 104 | + cin >> grid[i][j]; |
| 105 | + } |
| 106 | + } |
| 107 | + cout << maximumChocolates(r, c, grid); |
| 108 | +} |
| 109 | + |
| 110 | +// Sample Inputs |
| 111 | + |
| 112 | +// 3 4 |
| 113 | +// 2 3 1 2 |
| 114 | +// 3 4 2 2 |
| 115 | +// 5 6 3 5 |
| 116 | + |
| 117 | +// 2 2 |
| 118 | +// 1 1 |
| 119 | +// 1 2 |
| 120 | + |
| 121 | +// Corresponding Outputs |
| 122 | + |
| 123 | +// 21 |
| 124 | + |
| 125 | +// 5 |
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