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  • leetcode 1219. Path with Maximum Gold

    In a gold mine grid of size m * n, each cell in this mine has an integer representing the amount of gold in that cell, 0 if it is empty.

    Return the maximum amount of gold you can collect under the conditions:

    • Every time you are located in a cell you will collect all the gold in that cell.
    • From your position you can walk one step to the left, right, up or down.
    • You can't visit the same cell more than once.
    • Never visit a cell with 0 gold.
    • You can start and stop collecting gold from any position in the grid that has some gold.

    Example 1:

    Input: grid = [[0,6,0],[5,8,7],[0,9,0]]
    Output: 24
    Explanation:
    [[0,6,0],
     [5,8,7],
     [0,9,0]]
    Path to get the maximum gold, 9 -> 8 -> 7.
    

    Example 2:

    Input: grid = [[1,0,7],[2,0,6],[3,4,5],[0,3,0],[9,0,20]]
    Output: 28
    Explanation:
    [[1,0,7],
     [2,0,6],
     [3,4,5],
     [0,3,0],
     [9,0,20]]
    Path to get the maximum gold, 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7.

    Constraints:

    • 1 <= grid.length, grid[i].length <= 15
    • 0 <= grid[i][j] <= 100
    • There are at most 25 cells containing gold.

    思路:常规题(回溯法)

     1 class Solution {
     2 private:
     3     //int dx[4] = {-1, 0, 1, 0};
     4     int d[5] = {0, 1, 0, -1, 0};
     5 public:
     6     int getMaximumGold(vector<vector<int>>& grid) {
     7         int m = grid.size();
     8         if (m == 0)
     9             return 0;
    10         int n = grid[0].size();
    11         int maxGold = 0;
    12         for (int i = 0; i < m; ++i) {
    13             for (int j = 0; j < n; ++j) {
    14                 if (grid[i][j] > 0) {
    15                     maxGold = max(maxGold, dfs(grid, i, j));
    16                 }
    17             }
    18         }
    19         return maxGold;
    20     }
    21     int dfs(vector<vector<int>> &grid, int i, int j) {
    22         int result = 0;
    23         int temp = grid[i][j];
    24         grid[i][j] = 0; //将grid[i][j]设为0,使得深层的dfs(同一条路径)不会再遍历这个点
    25         for (int k = 0; k < 4; ++k) { //遍历四个方向
    26             int x = i + d[k], y = j + d[k + 1];
    27             if (x >= 0 && x < grid.size() && y >= 0 && y <  grid[0].size() && grid[x][y] != 0) {
    28                 result = max(result, dfs(grid, x, y));
    29             }
    30         }
    31         grid[i][j] = temp; //将grid[i][j]恢复,使得不同路径还能访问这个点
    32         return result + grid[i][j];
    33     }
    34 };
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  • 原文地址:https://www.cnblogs.com/qinduanyinghua/p/11896311.html
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