76 lines
3.0 KiB
C++
76 lines
3.0 KiB
C++
#include <iostream>
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#include <vector>
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#include <cstring>
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/**
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* 1463. Cherry Pickup II
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* You are given a rows x cols matrix grid representing a field of cherries where grid[i][j] represents the number of cherries that you can collect from the (i, j) cell.
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* You have two robots that can collect cherries for you:
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* Robot #1 is located at the top-left corner (0, 0), and
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* Robot #2 is located at the top-right corner (0, cols - 1).
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* Return the maximum number of cherries collection using both robots by following the rules below:
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* From a cell (i, j), robots can move to cell (i + 1, j - 1), (i + 1, j), or (i + 1, j + 1).
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* When any robot passes through a cell, It picks up all cherries, and the cell becomes an empty cell.
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* When both robots stay in the same cell, only one takes the cherries.
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* Both robots cannot move outside of the grid at any moment.
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* Both robots should reach the bottom row in grid.
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*
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* Input: grid = [[3,1,1],[2,5,1],[1,5,5],[2,1,1]]
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* Output: 24
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* Explanation: Path of robot #1 and #2 are described in color green and blue respectively.
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* Cherries taken by Robot #1, (3 + 2 + 5 + 2) = 12.
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* Cherries taken by Robot #2, (1 + 5 + 5 + 1) = 12.
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* Total of cherries: 12 + 12 = 24.
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*/
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class Solution {
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private:
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int dp[2][70][70];
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public:
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int cherryPickup(const std::vector<std::vector<int>>& grid) {
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int m = grid.size();
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int n = grid[0].size();
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// Set as -INF
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std::memset(this->dp, 0xC0, sizeof this->dp);
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this->dp[0][0][n - 1] = grid[0][0] + grid[0][n - 1];
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for (int i = 1; i < m; ++i) { // i <- row
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for (int j = 0; j < n; ++j) { // j <- col of robot 1
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for (int k = 0; k < n; ++k) { // k <- robot 2
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int thisPick = (j == k) ? (grid[i][j]) : (grid[i][j] + grid[i][k]);
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dp[i & 1][j][k] = std::max(dp[i & 1][j][k], dp[(i & 1) ^ 1][j][k] + thisPick);
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if (k > 0)
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dp[i & 1][j][k] = std::max(dp[i & 1][j][k], dp[(i & 1) ^ 1][j][k - 1] + thisPick);
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if (k < n - 1)
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dp[i & 1][j][k] = std::max(dp[i & 1][j][k], dp[(i & 1) ^ 1][j][k + 1] + thisPick);
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if (j > 0) {
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dp[i & 1][j][k] = std::max(dp[i & 1][j][k], dp[(i & 1) ^ 1][j - 1][k] + thisPick);
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if (k > 0)
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dp[i & 1][j][k] = std::max(dp[i & 1][j][k], dp[(i & 1) ^ 1][j - 1][k - 1] + thisPick);
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if (k < n - 1)
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dp[i & 1][j][k] = std::max(dp[i & 1][j][k], dp[(i & 1) ^ 1][j - 1][k + 1] + thisPick);
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}
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if (j < n - 1) {
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dp[i & 1][j][k] = std::max(dp[i & 1][j][k], dp[(i & 1) ^ 1][j + 1][k] + thisPick);
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if (k > 0)
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dp[i & 1][j][k] = std::max(dp[i & 1][j][k], dp[(i & 1) ^ 1][j + 1][k - 1] + thisPick);
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if (k < n - 1)
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dp[i & 1][j][k] = std::max(dp[i & 1][j][k], dp[(i & 1) ^ 1][j + 1][k + 1] + thisPick);
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}
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}
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}
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}
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int ret = -(0x7FFFFFFF) - 1;
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for (int i = 0; i < n; ++i)
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for (int j = 0; j < n; ++j)
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ret = std::max(ret, dp[(m ^ 1) & 1][i][j]);
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return ret;
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}
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};
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int main() {
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Solution solver{};
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auto ret = solver.cherryPickup({{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}});
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std::cout << ret;
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return 0;
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} |