代码随想录|图论|06岛屿数量（广搜BFS）

BFS版本

代码随想录|图论|05岛屿数量（深搜DFS）-CSDN博客

现在换成DFS来写，dfs的作用还是对当前点的四周进行标记！

#include <bits/stdc++.h>
using namespace std;// 定义四个方向
int dir[4][2] = {-1, 0, 0, 1, 1, 0, 0, -1};void bfs(vector<vector<int>> &grid, vector<vector<bool>> &visited, int x, int y)
{// 队列里面的是已经走过的节点queue<pair<int, int>> q;q.push({x, y});visited[x][y] = true;   // 只要加入队列，立刻标记// 直到队列为空，说明所有的点都被遍历了while (!q.empty()){pair<int, int> cur = q.front();q.pop();int curx = cur.first;int cury = cur.second;for (int i = 0; i < 4; i++){int nextx = curx + dir[i][0];int nexty = cury + dir[i][1];if (nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size())continue;if (grid[nextx][nexty] == 1 && !visited[nextx][nexty]){q.push({nextx, nexty});visited[nextx][nexty] = true;   // 只要加入队列，立刻标记}}}
}int main()
{int n, m;cin >> n >> m;vector<vector<int>> grid(n, vector<int>(m, 0));vector<vector<bool>> visited(n, vector<bool>(m, false));for (int i = 0; i < n; i++){for (int j = 0; j < m; j++){cin >> grid[i][j];}}int result = 0;for (int i = 0; i < n; i++){for (int j = 0; j < m; j++){if (grid[i][j] == 1 && !visited[i][j]){result++;bfs(grid, visited, i, j);}}}cout << result << endl;return 0;
}