专题十四——BFS

一BFS解决FloodFill算法

1图像渲染

2岛屿数量

3岛屿的最大面积

4被环绕的区域

二BFS解决蛋源最短路径问题

1迷宫中离入口最近的出口

2最小基因变化

3单词接龙

4为高尔夫比赛砍树

三BFS解决多源最短路径问题

1 01矩阵

2飞地的数量

3地图中的最高点

4地图分析

一BFS解决FloodFill算法

floodfill算法中文名：洪水灌溉：这类问题都是要来找出性质相同的联通块；解决好一道后，后面遇到类型的题目代码也是类似的！

1图像渲染

oj链接：图像渲染

解法：bfs

1.先把开始的坐标{sr，sc}放进队列中；

2.取出队头坐标{x，y}，直接将它改成color(队列坐标都是合法的，待改的)，再把它上下左右的坐标(等于image[sr][sc])放到队列中...

//dfs
class Solution
{
public:int target, newcolor, m, n;int dx[4] = {1, -1, 0, 0};int dy[4] = {0, 0, 1, -1};void dfs(vector<vector<int>> &image, int i, int j){image[i][j] = newcolor;for (int a = 0; a < 4; a++){int x = i + dx[a], y = j + dy[a];if (x >= 0 && x < m && y >= 0 && y < n && image[x][y] == target){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; // 特殊判断m = image.size(), n = image[0].size();target = image[sr][sc];newcolor = color;dfs(image, sr, sc);return image;}
};//bfs
class Solution
{
public:vector<vector<int>> floodFill(vector<vector<int>> &image, int sr, int sc, int color){if (image[sr][sc] == color)return image; // 特殊判断// bfs前准备int dx[4] = {1, -1, 0, 0};int dy[4] = {0, 0, 1, -1};int m = image.size(), n = image[0].size();int target = image[sr][sc];// bfs主逻辑queue<pair<int, int>> qe;qe.push({sr, sc});while (qe.size()){int sz = qe.size();for (int i = 0; i < sz; i++){auto &[a, b] = qe.front();qe.pop();if (image[a][b] == target)image[a][b] = color;for (int j = 0; j < 4; j++){int x = a + dx[j], y = b + dy[j];if (x >= 0 && x < m && y >= 0 && y < n && image[x][y] == target){qe.push({x, y});}}}}return image;}
};

2岛屿数量

oj链接：岛屿数量

解法：bfs

1. 两层for循环找‘1’&&该位置没被标记，将该坐标进队列，执行一次bfs；

2.bfs的主逻辑与上题类似：但在这里要注意：坐标进队列后要立刻马上进行标记！如果是取出节点后标记的话，会有重复节点进队列导致超时！！

//dfs
class Solution
{
public:bool vis[300][300] = {false};int dx[4] = {0, 0, 1, -1}, dy[4] = {1, -1, 0, 0};int m, n;void dfs(vector<vector<char>> &grid, int i, int j){vis[i][j] = true;for (int a = 0; a < 4; a++){int x = i + dx[a], y = j + dy[a];if (x >= 0 && x < m && y >= 0 && 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();int ret = 0;for (int i = 0; i < m; i++){for (int j = 0; j < n; j++){if (grid[i][j] == '1' && !vis[i][j]){ret++;dfs(grid, i, j);}}}return ret;}
};//bfs
class Solution
{
public:bool vis[300][300] = {false};int dx[4] = {0, 0, 1, -1}, dy[4] = {1, -1, 0, 0};int m, n;queue<pair<int, int>> q;void bfs(vector<vector<char>> &grid, int i, int j){//进队列后必须立刻进行标记！q.push({i, j});vis[i][j] = true;while (q.size()){auto [a, b] = q.front();q.pop();// vis[a][b]=true;节点会重复进队列，导致超时for (int k = 0; k < 4; k++){int x = a + dx[k], y = b + dy[k];if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == '1' && !vis[x][y]){q.push({x, y});vis[x][y] = true;}}}}int numIslands(vector<vector<char>> &grid){m = grid.size(), n = grid[0].size();int ret = 0;for (int i = 0; i < m; i++){for (int j = 0; j < n; j++){if (grid[i][j] == '1' && !vis[i][j]){ret++;bfs(grid, i, j);}}}return ret;}
};

3岛屿的最大面积

oj链接：岛屿的最大面积

与上题思路类似：只不过我们要求的是最大面积，我们可以：进行dfs后把岛屿的面积带出来

class Solution
{
public:int dx[4] = {-1, 1, 0, 0}, dy[4] = {0, 0, 1, -1};bool vis[50][50] = {false};int m, n;queue<pair<int, int>> q;void bfs(vector<vector<int>> &grid, int i, int j, int *area){q.push({i, j});vis[i][j] = true;while (q.size()){auto [a, b] = q.front();q.pop();for (int k = 0; k < 4; k++){int x = a + dx[k], y = b + dy[k];if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 1 && !vis[x][y]){q.push({x, y});vis[x][y] = true;(*area)++;}}}}int maxAreaOfIsland(vector<vector<int>> &grid){m = grid.size(), n = grid[0].size();int MaxArea = 0;for (int i = 0; i < m; i++){for (int j = 0; j < n; j++){if (grid[i][j] == 1 && !vis[i][j]){int area = 1; // 从该位置进行bfs：找出该位置处岛屿的面积bfs(grid, i, j, &area);MaxArea = max(MaxArea, area);}}}return MaxArea;}
};

4被环绕的区域

oj链接：被围绕的区域

解法：bfs

此题在递归搜索与回溯算法题

class Solution
{
public:bool vis[200][200];int dx[4] = {1, -1, 0, 0}, dy[4] = {0, 0, 1, -1};int m, n;queue<pair<int, int>> q;void bfs(vector<vector<char>> &board, int i, int j){q.push({i, j});vis[i][j] = true;while (q.size()){auto [a, b] = q.front();q.pop();for (int k = 0; k < 4; k++){int x = a + dx[k], y = b + dy[k];if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'O' && !vis[x][y]){q.push({x, y});vis[x][y] = true;}}}}void solve(vector<vector<char>> &board){m = board.size(), n = board[0].size();for (int i = 0; i < m; i++){if (board[i][0] == 'O')bfs(board, i, 0);if (board[i][n - 1] == 'O')bfs(board, i, n - 1);}for (int j = 0; j < n; j++){if (board[0][j] == 'O')bfs(board, 0, j);if (board[m - 1][j] == 'O')bfs(board, m - 1, j);}for (int i = 0; i < m; i++){for (int j = 0; j < n; j++){if (board[i][j] == 'O' && !vis[i][j])board[i][j] = 'X';}}}
};

二BFS解决蛋源最短路径问题

1迷宫中离入口最近的出口

oj链接：迷宫中离入口最近的出口

从开始粗来一次bfs遍历即可；如果进队列的坐标是出口，直接返回遍历的层数ret即可！

class Solution
{
public:bool vis[100][100];int dx[4] = {0, 0, 1, -1}, dy[4] = {1, -1, 0, 0};int m, n;int nearestExit(vector<vector<char>> &maze, vector<int> &entrance){m = maze.size(), n = maze[0].size();queue<pair<int, int>> q;int beginx = entrance[0], beginy = entrance[1];q.push({beginx, beginy});vis[beginx][beginy] = true;int ret = 0;while (q.size()){ret++; // 往外扩int sz = q.size();for (int i = 0; i < sz; i++){auto [a, b] = q.front();q.pop();for (int k = 0; k < 4; k++){int x = a + dx[k], y = b + dy[k];if (x >= 0 && x < m && y >= 0 && y < n && maze[x][y] == '.' &&!vis[x][y]){if (x == 0 || x == m - 1 || y == 0 || y == n - 1)return ret; // 找到出口了，直接返回遍历的层数,即最短路径q.push({x, y});vis[x][y] = true;}}}}return -1;}
};

2最小基因变化

oj链接：最小基因变化

class Solution
{
public:int minMutation(string startGene, string endGene, vector<string> &bank){char arr[4] = {'A', 'G', 'C', 'T'};unordered_map<string, bool> vis;for (auto &s : bank)vis[s] = false;if(!vis.count(endGene)) return -1;// 不存在直接返回queue<string> q;q.push(startGene);vis[startGene] = true; // 有可能基因库也有这个，提前标记int ret = 0;while (q.size()){ret++;int sz = q.size();for (int i = 0; i < sz; i++){string s = q.front();q.pop();for (int i = 0; i < 8; i++){for (int j = 0; j < 4; j++){string tmp = s;tmp[i] = arr[j]; // 这里就没必要判断tmp[i]是否与arr[j]相等:因为前面我们已经标记过了if (vis.count(tmp) && !vis[tmp]){if (tmp == endGene)return ret;q.push(tmp);vis[tmp] = true;}}}}}return -1;}
};

3单词接龙

oj链接：单词接龙

与上题思路类似

class Solution
{
public:int ladderLength(string beginWord, string endWord, vector<string> &wordList){string t = "qwertyuiopasdfghjklzxcvbnm";unordered_map<string, bool> vis;for (auto &word : wordList)vis[word] = false;if (!vis.count(endWord))return 0;int ans = 1;queue<string> qe;qe.push(beginWord);vis[beginWord] = true;while (qe.size()){ans++;int sz = qe.size();while (sz--){string s = qe.front();qe.pop();for (int i = 0; i < s.size(); i++){char ch = s[i];for (int j = 0; j < 26; j++){s[i] = t[j];if (vis.count(s) && !vis[s]){if (s == endWord)return ans;qe.push(s);vis[s] = true;}}s[i] = ch;}}}return 0;}
};

4为高尔夫比赛砍树

oj链接：为高尔夫比赛砍树

注意：

不是说一定要砍，当初做的时候就是看劈叉了~

class Solution
{
public:typedef pair<int, int> PII;int n, m;int cutOffTree(vector<vector<int>> &forest){// 先预处理砍树顺序vector<PII> trees;n = forest.size(), m = forest[0].size();for (int i = 0; i < n; i++){for (int j = 0; j < m; j++){if (forest[i][j] > 1)trees.push_back({i, j});}}sort(trees.begin(), trees.end(), [&](const PII &p1, const PII &p2){ return forest[p1.first][p1.second] < forest[p2.first][p2.second]; });// 若干个迷宫问题的最小步数累加int dx = 0, dy = 0;int ret = 0;for (auto &[a, b] : trees){int step = bfs(forest, dx, dy, a, b);if (step == -1)return -1;ret += step;dx = a, dy = b; // 更新位置}return ret;}int dx[4] = {0, 0, 1, -1};int dy[4] = {1, -1, 0, 0};int bfs(vector<vector<int>> &forest, int bx, int by, int ex, int ey){if (bx == ex && by == ey)return 0;int step = 0;queue<PII> qe;bool vis[50][50] = {false};// 层序遍历主逻辑qe.push({bx, by});vis[bx][by] = true;while (qe.size()){step++;int size = qe.size();while (size--){auto [a, b] = qe.front();qe.pop();for (int i = 0; i < 4; i++){int x = dx[i] + a, y = dy[i] + b;if (x >= 0 && x < n && y >= 0 && y < m && !vis[x][y] && forest[x][y] != 0){if (x == ex && y == ey)return step;vis[x][y] = true;qe.push({x, y});}}}}return -1;}
};

三BFS解决多源最短路径问题

1 01矩阵

oj链接：01 矩阵

解法：BFS + 正难则反

如果从1作为起点走到终点0，那等到终点时记录距离时就不知道是从哪一个1为起点的

所以我们从0为起点开始，走到不是0的位置就进行距离记录，同时把它

class Solution
{
public:// bool vis[1000][1000];//开10000会溢出?int dx[4] = {0, 0, 1, -1}, dy[4] = {1, -1, 0, 0};int m, n;vector<vector<int>> updateMatrix(vector<vector<int>> &mat){m = mat.size(), n = mat[0].size();vector<vector<int>> dist(m, vector<int>(n, -1));queue<pair<int, int>> q;for (int i = 0; i < m; i++){for (int j = 0; j < n; j++){if (mat[i][j] == 0){dist[i][j] = 0;q.push({i, j});}}}// int step=0;while (q.size()){// step++;// int sz=q.size();// while(sz--)// {auto [a, b] = q.front();q.pop();for (int i = 0; i < 4; i++){int x = dx[i] + a, y = dy[i] + b;// if(x>=0&&x<m&&y>=0&&y<n&&!vis[x][y])if (x >= 0 && x < m && y >= 0 && y < n && dist[x][y] == -1){// dist[x][y]=cnt;// vis[x][y]=true;dist[x][y] = dist[a][b] + 1;q.push({x, y});}}// }}return dist;}
};

2飞地的数量

oj链接：飞地的数量

解法：dfs + 正难则反

a先把边界上的1进入队列并进行标记；

b进行dfs遍历(上下左右找合法坐标)只要它是1并且没被标记，就把它加入到队列中并进行标记；

c再次对grid进行遍历，只要坐标的值是1并且没别标记就代表它走不出边界，进行统计即可

class Solution
{
public:bool vis[500][500];int dx[4] = {0, 0, 1, -1}, dy[4] = {1, -1, 0, 0};int m, n;int numEnclaves(vector<vector<int>> &grid){m = grid.size(), n = grid[0].size();queue<pair<int, int>> q;// 边上1统计for (int i = 0; i < m; i++){for (int j = 0; j < n; j++){if (i == 0 || i == m - 1 || j == 0 || j == n - 1){if (grid[i][j] == 1){q.push({i, j});vis[i][j] = true;}}}}// bfs进行遍历while (q.size()){// int sz=q.size();// while(sz--)// {auto [a, b] = q.front();q.pop();for (int k = 0; k < 4; k++){int x = dx[k] + a, y = dy[k] + b;if (x >= 0 && x < m && y >= 0 && y < n && !vis[x][y] && grid[x][y] == 1){q.push({x, y});vis[x][y] = true;}}// }}int ret = 0;for (int i = 0; i < m; i++){for (int j = 0; j < n; j++){if (!vis[i][j] && grid[i][j] == 1)ret++;}}return ret;}
};

3地图中的最高点

oj链接：地图中的最高点

解法：bfs

安排高度时，任意相邻的格子高度差至多为1，也就是既可以设计成0，也可以设计成1；

但题目要我们返回的是高度值最大的情况，所以我们贪心地把所有高度差都设计成1即可；

也就是从所有水域位置为起点来来依次bfs遍历即可完成任务~

class Solution {
public:typedef pair<int,int> PII;int dx[4]={0,0,1,-1};int dy[4]={1,-1,0,0};vector<vector<int>> highestPeak(vector<vector<int>>& isWater) {queue<PII> qe;int n=isWater.size(),m=isWater[0].size();vector<vector<int>> hight(n,vector<int>(m,-1));for(int i=0;i<n;i++){for(int j=0;j<m;j++){if(isWater[i][j]==1){qe.push({i,j});hight[i][j]=0;}}}while(qe.size()){int sz=qe.size();while(sz--){auto[a,b]=qe.front();qe.pop();for(int i=0;i<4;i++){int x=dx[i]+a,y=dy[i]+b;if(x>=0&&x<n&&y>=0&&y<m&&hight[x][y]==-1){hight[x][y]=hight[a][b]+1;qe.push({x,y});}}}}return hight;}
};

4地图分析

oj链接：地图分析

与01矩阵的思路一样~

//用dist数组
class Solution
{
public:typedef pair<int, int> PII;int dx[4] = {0, 0, 1, -1};int dy[4] = {1, -1, 0, 0};int maxDistance(vector<vector<int>> &grid){int n = grid.size(), m = grid[0].size();vector<vector<int>> dist(n, vector<int>(m, -1));queue<PII> qe;for (int i = 0; i < n; i++){for (int j = 0; j < m; j++){if (grid[i][j] == 1){qe.push({i, j});dist[i][j] = 0;}}}if (qe.size() == 0 || qe.size() == n * m)return -1; // 边界情况int ans = 0;while (qe.size()){auto [a, b] = qe.front();qe.pop();for (int i = 0; i < 4; i++){int x = dx[i] + a, y = dy[i] + b;if (x >= 0 && x < n && y >= 0 && y < m && dist[x][y] == -1){dist[x][y] = dist[a][b] + 1;qe.push({x, y});ans = max(ans, dist[x][y]);}}}return ans;}
};//统计遍历层数
class Solution
{
public:bool vis[100][100];int dx[4] = {1, -1, 0, 0}, dy[4] = {0, 0, 1, -1};int m, n, ret = -1;int maxDistance(vector<vector<int>> &grid){m = grid.size(), n = grid[0].size();queue<pair<int, int>> q;for (int i = 0; i < m; i++){for (int j = 0; j < n; j++){if (grid[i][j] == 1){q.push({i, j});vis[i][j] = true;}}}if (q.size() == 0 || q.size() == m * n)return ret;while (q.size()){ret++;int sz = q.size();while (sz--){auto [a, b] = q.front();q.pop();for (int i = 0; i < 4; i++){int x = a + dx[i], y = b + dy[i];if (x >= 0 && x < m && y >= 0 && y < n && !vis[x][y]){q.push({x, y});vis[x][y] = true;}}}}return ret;}
};