1219.黄金矿工
你要开发一座金矿,地质勘测学家已经探明了这座金矿中的资源分布,并用大小为 m * n 的网格 grid 进行了标注。每个单元格中的整数就表示这一单元格中的黄金数量;如果该单元格是空的,那么就是 0。
为了使收益最大化,矿工需要按以下规则来开采黄金:
每当矿工进入一个单元,就会收集该单元格中的所有黄金。
矿工每次可以从当前位置向上下左右四个方向走。
每个单元格只能被开采(进入)一次。
不得开采(进入)黄金数目为 0 的单元格。
矿工可以从网格中 任意一个 有黄金的单元格出发或者是停止。
示例 1:
输入:grid = [[0,6,0],[5,8,7],[0,9,0]]
输出:24
解释:
[[0,6,0],
[5,8,7],
[0,9,0]]
一种收集最多黄金的路线是:9 -> 8 -> 7。
示例 2:
输入:grid = [[1,0,7],[2,0,6],[3,4,5],[0,3,0],[9,0,20]]
输出:28
解释:
[[1,0,7],
[2,0,6],
[3,4,5],
[0,3,0],
[9,0,20]]
一种收集最多黄金的路线是:1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7。
提示:
1 <= grid.length, grid[i].length <= 15
0 <= grid[i][j] <= 100
最多 25 个单元格中有黄金。
一开始我就是想的递归加回溯
但是我嫌麻烦,就用了值传递
class Solution { private: vector<vector<int>> _grid; int getGold(int i, int j, vector<vector<int>> flagCache, int totalGold) { if(i<0||j<0) return totalGold; if(i>=_grid.size()) return totalGold; if(j>=_grid[i].size()) return totalGold; if(_grid[i][j] == 0||flagCache[i][j] == 1) return totalGold; flagCache[i][j] = 1; vector<int> goldNumCache; goldNumCache.push_back(getGold(i-1,j,flagCache,totalGold + _grid[i][j])); goldNumCache.push_back(getGold(i+1,j,flagCache,totalGold + _grid[i][j])); goldNumCache.push_back(getGold(i,j-1,flagCache,totalGold + _grid[i][j])); goldNumCache.push_back(getGold(i,j+1,flagCache,totalGold + _grid[i][j])); sort(goldNumCache.begin(),goldNumCache.end()); return goldNumCache.back(); } public: int getMaximumGold(vector<vector<int>>& grid) { _grid = grid; vector<vector<int>> _flag = vector<vector<int>>(grid.size()); for(int i=0;i<grid.size();i++) _flag[i] = vector<int>(grid[i].size()); vector<int> goldNum; for(int i=0;i<grid.size();i++) { for(int j=0;j<grid[i].size();j++) { goldNum.push_back(getGold(i,j,_flag,0)); } } sort(goldNum.begin(),goldNum.end()); return goldNum.back(); } };
改成引用传递之后我本来还想着在递归调用函数之后回溯的
int getGold(int i, int j, int totalGold) { if(i<0||j<0) return totalGold; if(i>=_grid.size()) return totalGold; if(j>=_grid[i].size()) return totalGold; if(_grid[i][j] == 0||_flag[i][j] == 1) return totalGold; _flag[i][j] = 1; vector<int> goldNumCache; // 递归 goldNumCache.push_back(getGold(i-1,j,totalGold + _grid[i][j])); // 回溯 if(goldNumCache.back() != totalGold + _grid[i][j]) _flag[i-1][j] = 0; // 递归 goldNumCache.push_back(getGold(i+1,j,totalGold + _grid[i][j])); // 回溯 if(goldNumCache.back() != totalGold + _grid[i][j]) _flag[i+1][j] = 0; // 递归 goldNumCache.push_back(getGold(i,j-1,totalGold + _grid[i][j])); // 回溯 if(goldNumCache.back() != totalGold + _grid[i][j]) _flag[i][j-1] = 0; // 递归 goldNumCache.push_back(getGold(i,j+1,totalGold + _grid[i][j])); // 回溯 if(goldNumCache.back() != totalGold + _grid[i][j]) _flag[i][j+1] = 0; sort(goldNumCache.begin(),goldNumCache.end()); return goldNumCache.back(); }
然后这样就错了……
我还在想哪里错了……但是我突然发现其实可以不用想这个了,它本身就麻烦
本来最简单的应该是在返回函数的时候回溯
就,只管自己的状态,这个原则,要遵守
而且确实可以允许这样做,我回溯在递归调用之后就是合理的
int getGold(int i, int j, int totalGold) { if(i<0||j<0) return totalGold; if(i>=_grid.size()) return totalGold; if(j>=_grid[i].size()) return totalGold; if(_grid[i][j] == 0||_flag[i][j] == 1) return totalGold; _flag[i][j] = 1; vector<int> goldNumCache; // 递归 goldNumCache.push_back(getGold(i-1,j,totalGold + _grid[i][j])); goldNumCache.push_back(getGold(i+1,j,totalGold + _grid[i][j])); goldNumCache.push_back(getGold(i,j-1,totalGold + _grid[i][j])); goldNumCache.push_back(getGold(i,j+1,totalGold + _grid[i][j])); sort(goldNumCache.begin(),goldNumCache.end()); _flag[i][j] = 0; return goldNumCache.back(); }
最终代码:
class Solution { private: vector<vector<int>> _grid; vector<vector<int>> _flag; int getGold(int i, int j, int totalGold) { if(i<0||j<0) return totalGold; if(i>=_grid.size()) return totalGold; if(j>=_grid[i].size()) return totalGold; if(_grid[i][j] == 0||_flag[i][j] == 1) return totalGold; _flag[i][j] = 1; vector<int> goldNumCache; // 递归 goldNumCache.push_back(getGold(i-1,j,totalGold + _grid[i][j])); goldNumCache.push_back(getGold(i+1,j,totalGold + _grid[i][j])); goldNumCache.push_back(getGold(i,j-1,totalGold + _grid[i][j])); goldNumCache.push_back(getGold(i,j+1,totalGold + _grid[i][j])); sort(goldNumCache.begin(),goldNumCache.end()); _flag[i][j] = 0; return goldNumCache.back(); } public: int getMaximumGold(vector<vector<int>>& grid) { _grid = grid; _flag = vector<vector<int>>(grid.size()); for(int i=0;i<grid.size();i++) _flag[i] = vector<int>(grid[i].size()); vector<int> goldNum; for(int i=0;i<grid.size();i++) { for(int j=0;j<grid[i].size();j++) { goldNum.push_back(getGold(i,j,0)); } } sort(goldNum.begin(),goldNum.end()); return goldNum.back(); } };
还是超时了……不懂为啥……
看了一下别人的题解,感觉跟我的一样啊
别人不用 flag,用 grid 置 0 代替,我试试
class Solution { private: vector<vector<int>> _grid; int getGold(int i, int j, int totalGold) { if(i<0||j<0) return totalGold; if(i>=_grid.size()) return totalGold; if(j>=_grid[i].size()) return totalGold; if(_grid[i][j] == 0) return totalGold; int goldCache = _grid[i][j]; _grid[i][j] = 0; vector<int> goldNumCache; // 递归 goldNumCache.push_back(getGold(i-1,j,totalGold + goldCache)); goldNumCache.push_back(getGold(i+1,j,totalGold + goldCache)); goldNumCache.push_back(getGold(i,j-1,totalGold + goldCache)); goldNumCache.push_back(getGold(i,j+1,totalGold + goldCache)); sort(goldNumCache.begin(),goldNumCache.end()); _grid[i][j] = goldCache; return goldNumCache.back(); } public: int getMaximumGold(vector<vector<int>>& grid) { _grid = grid; vector<int> goldNum; for(int i=0;i<grid.size();i++) { for(int j=0;j<grid[i].size();j++) { goldNum.push_back(getGold(i,j,0)); } } sort(goldNum.begin(),goldNum.end()); return goldNum.back(); } };
还是超时了……思考……
让我再看看别人的题解
别人是先判断接下来要走的格子能不能走,然后再调用函数
我是先调用函数然后再判断自己的状态
我记得好像是调用函数也是要耗费时间的,所以才有内联这个东西
再让我试试
class Solution { private: static constexpr int delta[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}}; public: int getMaximumGold(vector<vector<int>>& grid) { function<int(int,int,int)> getGold = [&](int x, int y, int totalGold) { vector<int> totalGoldCache; int goldCache = grid[x][y]; grid[x][y] = 0; // 递归 for(int i=0;i<4;i++) { int nx = x + delta[i][0]; int ny = y + delta[i][1]; if((nx >= 0) && (ny >= 0) && (nx < grid.size())) if(ny < grid[nx].size()) if(grid[nx][ny] != 0) totalGoldCache.push_back(getGold(nx, ny, totalGold + goldCache)); } sort(totalGoldCache.begin(), totalGoldCache.end()); grid[x][y] = goldCache; return totalGoldCache.back(); }; vector<int> goldNum; for(int i=0;i<grid.size();i++) { for(int j=0;j<grid[i].size();j++) { if(grid[i][j] != 0) goldNum.push_back(getGold(i,j,0)); } } sort(goldNum.begin(),goldNum.end()); return goldNum.back(); } };
这个会出错……因为取 vector 的 back 的时候不一定有值
还需要预先加上一个默认的值
class Solution { private: static constexpr int delta[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}}; public: int getMaximumGold(vector<vector<int>>& grid) { int m = grid.size(), n = grid[0].size(); function<int(int,int,int)> getGold = [&](int x, int y, int totalGold) { vector<int> totalGoldCache; int goldCache = grid[x][y]; grid[x][y] = 0; // 递归 for(int i=0;i<4;i++) { int nx = x + delta[i][0]; int ny = y + delta[i][1]; if (nx >= 0 && nx < m && ny >= 0 && ny < n && grid[nx][ny] > 0) totalGoldCache.push_back(getGold(nx, ny, totalGold + goldCache)); } totalGoldCache.push_back(totalGold + goldCache); sort(totalGoldCache.begin(), totalGoldCache.end()); grid[x][y] = goldCache; return totalGoldCache.back(); }; vector<int> totalGoldCache2; for(int i=0;i<grid.size();i++) { for(int j=0;j<grid[i].size();j++) { if(grid[i][j] != 0) totalGoldCache2.push_back(getGold(i,j,0)); } } totalGoldCache2.push_back(0); sort(totalGoldCache2.begin(),totalGoldCache2.end()); return totalGoldCache2.back(); } };
这样就好了