【算法】网络图中的dfs-编程知识

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文章目录

引言
一、单词搜索
二、黄金矿工
三、不同路径 |||
四、图像渲染
五、岛屿数量
六、岛屿的最大面积
七、被围绕的区域
八、太平洋大西洋水流问题
九、扫雷游戏
总结

引言

在二维网络图中的dfs，反而一般不需要画决策树，因为在二维图像中有时候很直观可以看出决策关系，一般为上下左右搜索。

一、单词搜索

细节：

dfs函数设置返回值bool，以便及时调整路线
设置向量数组dx，dy
设置bool数组vis，实现剪枝
pos设置为函数参数，方便回溯

class Solution
{int dx[4] = {1, -1, 0, 0};int dy[4] = {0, 0, 1, -1};bool vis[7][7];int m, n;
public:bool dfs(vector<vector<char>>& board, int i, int j, string& word, int pos){if(pos == word.size()) return true;vis[i][j] = true;for(int k=0; k<4; ++k){int x = i + dx[k], y = j + dy[k];if(x >= 0 && y >= 0 && x < m && y < n && !vis[x][y] && board[x][y] == word[pos]){if(dfs(board, x, y, word, pos + 1)) return true;}}vis[i][j] = false;return false;}bool exist(vector<vector<char>>& board, string& word){m = board.size(), n = board[0].size();for(int i=0; i<m; ++i){for(int j=0; j<n; ++j){if(board[i][j] == word[0]){if(dfs(board, i, j, word, 1)) return true;}}}return false;}
};

二、黄金矿工

细节：

设置向量数组dx，dy
设置bool数组vis，实现剪枝
sum设置为函数参数，方便回溯

class Solution
{int ret = 0;bool vis[16][16];int dx[4] = {1, -1, 0, 0};int dy[4] = {0, 0, 1, -1};int m, n;
public:void dfs(vector<vector<int>>& grid, int i, int j, int sum){vis[i][j] = true;for(int k=0; k<4; ++k){int x = i + dx[k], y = j + dy[k];if(x >= 0 && y >= 0 && x < m && y < n&& !vis[x][y] && grid[x][y]){dfs(grid, x, y, sum + grid[x][y]);}else ret = max(ret, sum);}vis[i][j] = false;}int getMaximumGold(vector<vector<int>>& grid){m = grid.size(), n = grid[0].size();for(int i=0; i<m; ++i){for(int j=0; j<n; ++j){if(grid[i][j]){dfs(grid, i, j, grid[i][j]);}}}return ret;}
};

三、不同路径 |||

细节：

设置向量数组dx，dy
设置bool数组vis，实现剪枝
path设置为函数参数，方便回溯

class Solution
{int ret = 0;int dx[4] = {1, -1, 0, 0};int dy[4] = {0, 0, 1, -1};bool vis[20][20];int m, n, count = 0;
public:void dfs(vector<vector<int>>& grid, int i, int j, int path){if(grid[i][j] == 2){if(path + count == m * n) ++ret;return;}vis[i][j] = true;for(int k=0; k<4; ++k){int x = i + dx[k], y = j + dy[k];if(x >= 0 && y >= 0 && x < m && y < n&& !vis[x][y] && grid[x][y] != -1){dfs(grid, x, y, path + 1);}}vis[i][j] = false;}int uniquePathsIII(vector<vector<int>>& grid){m = grid.size(), n = grid[0].size();int si, sj;for(int i=0; i<m; ++i){for(int j=0; j<n; ++j){if(grid[i][j] == -1) ++count;else if(grid[i][j] == 1) {si = i, sj = j;}}}dfs(grid, si, sj, 1);return ret;}
};

本题开始往后，是floodfill算法的练习
floodfill算法，本质就是寻找连通块
同时floodfill算法没有回溯

四、图像渲染

细节：

设置向量数组dx，dy
记录baseColor, newColor

class Solution
{int dx[4] = {1, -1, 0, 0};int dy[4] = {0, 0, 1, -1};int baseColor, newColor;int m, n;
public:void dfs(vector<vector<int>>& image, int i, int j){image[i][j] = newColor;for(int k=0; k<4; ++k){int x = i + dx[k], y = j + dy[k];if(x >= 0 && y >= 0 && x < m && y < n && image[x][y] == baseColor){dfs(image, x, y);}}}vector<vector<int>> floodFill(vector<vector<int>>& image, int sr, int sc, int color){if(image[sr][sc] == color) return image;baseColor = image[sr][sc], newColor = color;m = image.size(), n = image[0].size();dfs(image, sr, sc);return image;}
};

五、岛屿数量

细节：

设置向量数组dx，dy
设置bool数组vis，实现剪枝

class Solution
{int ret;bool vis[301][301];int dx[4] = {1, -1, 0, 0};int dy[4] = {0, 0, 1, -1};int m, n;
public:void dfs(vector<vector<char>>& grid, int i, int j){vis[i][j] = true;for(int k=0; k<4; ++k){int x = i + dx[k], y = j + dy[k];if(x >= 0 && y >= 0 && x < m && y < n&& !vis[x][y] && grid[x][y] == '1'){dfs(grid, x, y);}}}int numIslands(vector<vector<char>>& grid){m = grid.size(), n = grid[0].size();for(int i=0; i<m; ++i){for(int j=0; j<n; ++j){if(!vis[i][j] && grid[i][j] == '1'){dfs(grid, i, j);++ret;}}}return ret;}
};

六、岛屿的最大面积

细节：

设置向量数组dx，dy
设置bool数组vis，实现剪枝

class Solution
{int ret, sum;vector<vector<bool>> vis;int dx[4] = {1, -1, 0, 0};int dy[4] = {0, 0, 1, -1};int m, n;
public:void dfs(vector<vector<int>>& grid, int i, int j){++sum;vis[i][j] = true;for(int k=0; k<4; ++k){int x = i + dx[k], y = j + dy[k];if(x >= 0 && y >= 0 && x < m && y < n&& !vis[x][y] && grid[x][y] == 1){dfs(grid, x, y);}else ret = max(ret, sum);}}int maxAreaOfIsland(vector<vector<int>>& grid){m = grid.size(), n = grid[0].size();vis.resize(m, vector<bool>(n));for(int i=0; i<m; ++i){for(int j=0; j<n; ++j){if(!vis[i][j] && grid[i][j] == 1){sum = 0;dfs(grid, i, j);}}}return ret;}
};

七、被围绕的区域

细节：

正难则反，将边界上及其相连的‘O’先全部标记起来，最后遍历矩阵，将未被标记的‘O’改成‘X’
设置向量数组dx，dy
设置bool数组vis，实现剪枝

class Solution
{vector<vector<bool>> vis;int dx[4] = {1, -1, 0, 0};int dy[4] = {0, 0, 1, -1};int m, n;
public:void dfs(vector<vector<char>>& board, int i, int j){vis[i][j] = true;for(int k=0; k<4; ++k){int x = i + dx[k], y = j + dy[k];if(x >= 0 && y >= 0 && x < m && y < n&& !vis[x][y] && board[x][y] == 'O'){dfs(board, x, y);}}}void solve(vector<vector<char>>& board){m = board.size(), n = board[0].size();vis.resize(m, vector<bool>(n));for(int i=0; i<m; ++i){if(board[i][0] == 'O') dfs(board, i, 0);if(board[i][n-1] == 'O') dfs(board, i, n-1);}for(int j=0; j<n; ++j){if(board[0][j] == 'O') dfs(board, 0, j);if(board[m-1][j] == 'O') dfs(board, m-1, j);}for(int i=0; i<m; ++i){for(int j=0; j<n; ++j){if(!vis[i][j] && board[i][j] == 'O'){board[i][j] = 'X';}}}}
};

八、太平洋大西洋水流问题

思路：正难则反，水往高处流。

利用两个标记数组，分别记录太平洋和大西洋的水可以流到的区域，找出标记数组重叠的部分即可。
为了只用写一份dfs函数，将vis标记数组以参数的形式传递
设置向量数组dx，dy

class Solution
{int dx[4] = {1, -1, 0, 0};int dy[4] = {0, 0, 1, -1};int m, n;
public:void dfs(vector<vector<int>>& heights, int i, int j, vector<vector<bool>>& vis){vis[i][j] = true;for(int k=0; k<4; ++k){int x = i + dx[k], y = j + dy[k];if(x >= 0 && y >= 0 && x < m && y < n&& !vis[x][y] && heights[x][y] >= heights[i][j])//正难则反，水往高处流{dfs(heights, x, y, vis);}}}vector<vector<int>> pacificAtlantic(vector<vector<int>>& heights){m = heights.size(), n = heights[0].size();vector<vector<bool>> vis1(m, vector<bool>(n));vector<vector<bool>> vis2(m, vector<bool>(n));for(int i=0; i<m; ++i){if(!vis1[i][0]) dfs(heights, i, 0, vis1);if(!vis2[i][n-1]) dfs(heights, i, n-1, vis2);}for(int j=0; j<n; ++j){if(!vis1[0][j]) dfs(heights, 0, j, vis1);if(!vis2[m-1][j]) dfs(heights, m-1, j, vis2);}vector<vector<int>> ret;for(int i=0; i<m; ++i){for(int j=0; j<n; ++j){if(vis1[i][j] && vis2[i][j]){ret.push_back({i, j});}}}return ret;}
};

九、扫雷游戏

细节：本题实际上是一道模拟题，关键是理解题意。

点击位置是雷，直接改变返回
不是雷，则进入dfs函数递归
对于当前格子，先计算周围8个格子中雷的数量
- 如果有雷，将当前格子改为雷的数量，返回
- 如果无雷，将当前格子改为挖出的空方块，继续递归
递归的时候，要进行8个方向的搜索，所以向量数组要存储8个方向

class Solution
{bool vis[51][51];int dx[8] = {1, 1, 1, 0, 0, -1, -1, -1};int dy[8] = {1, 0, -1, 1, -1, 1, 0, -1};int m, n;
public:void dfs(vector<vector<char>>& board, int i, int j){if(board[i][j] == 'M') return;int count = 0;for(int k=0; k<8; ++k){int x = i + dx[k], y = j + dy[k];if(x >= 0 && y >= 0 && x < m && y < n && board[x][y] == 'M'){++count;}}if(count){board[i][j] = count + '0';return;}else board[i][j] = 'B';vis[i][j] = true;for(int k=0; k<8; ++k){int x = i + dx[k], y = j + dy[k];if(x >= 0 && y >= 0 && x < m && y < n && !vis[x][y]){dfs(board, x, y);}}}vector<vector<char>> updateBoard(vector<vector<char>>& board, vector<int>& click){int ci = click[0], cj = click[1];if(board[ci][cj] == 'M'){board[ci][cj] = 'X';return board;}m = board.size(), n = board[0].size();dfs(board, ci, cj);return board;}
};