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.

思路： 找最长路的过程是简单的遍历四个方向做DFS。数据规模比较小，可以直接遍历矩阵，找到不是0的点作为起点做DFS。在每次经过一个节点的时候，把它暂时置为0，表示不能再次经过它，等到递归结束的时候再改成原来的值即可。

class Solution {
public:
    int directions[5]={0,1,0,-1,0};
    int dfs(int i,int j,vector<vector<int>>& grid){
        int tem=grid[i][j];
        int x,y,result=0;
        grid[i][j]=0;
        for(int d=0;d<4;d++){
            x=i+directions[d];y=j+directions[d+1];
            if(x>=0&&y>=0&&x<grid.size()&&y<grid[0].size()&&grid[x][y])
                result=max(grid[x][y]+dfs(x,y,grid),result);
        }
        grid[i][j]=tem;
        return result;
    }
    int getMaximumGold(vector<vector<int>>& grid) {
        int result=0;
        for(int i=0;i<grid.size();i++){
            for(int j=0;j<grid[0].size();j++){
                if(grid[i][j]>0)
                    result=max(result,grid[i][j]+dfs(i,j,grid));
            }
        }
        return result;
        
    }
};