【代码随想录算法训练营第37期第五十七天 | 岛屿数量深搜、岛屿数量广搜、岛屿的最大面积】

时间:2025/8/4 8:05:41来源：https://blog.csdn.net/cattte/article/details/140224525 浏览次数: 0次

代码随想录算法训练营第37期第五十七天 | 岛屿数量深搜、岛屿数量广搜、岛屿的最大面积

一、岛屿数量深搜

解题代码C++：

#include <iostream>
#include <vector>
using namespace std;int dir[4][2] = {0, 1, 1, 0, -1, 0, 0, -1}; // 四个方向
void dfs(const vector<vector<int>>& grid, vector<vector<bool>>& visited, int x, int y) {for (int i = 0; i < 4; i++) {int nextx = x + dir[i][0];int nexty = y + dir[i][1];if (nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size()) continue;  // 越界了，直接跳过if (!visited[nextx][nexty] && grid[nextx][nexty] == 1) { // 没有访问过的 同时 是陆地的visited[nextx][nexty] = true;dfs(grid, visited, nextx, nexty);}}
}int main() {int n, m;cin >> n >> m;vector<vector<int>> grid(n, vector<int>(m, 0));for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {cin >> grid[i][j];}}vector<vector<bool>> visited(n, vector<bool>(m, false));int result = 0;for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {if (!visited[i][j] && grid[i][j] == 1) {visited[i][j] = true;result++; // 遇到没访问过的陆地，+1dfs(grid, visited, i, j); // 将与其链接的陆地都标记上 true}}}cout << result << endl;
}

题目链接/文章讲解/视频讲解：
https://www.programmercarl.com/kamacoder/0099.%E5%B2%9B%E5%B1%BF%E7%9A%84%E6%95%B0%E9%87%8F%E6%B7%B1%E6%90%9C.html

二、岛屿数量广搜

解题代码C++：

#include <iostream>
#include <vector>
#include <queue>
using namespace std;int dir[4][2] = {0, 1, 1, 0, -1, 0, 0, -1}; // 四个方向
void bfs(const vector<vector<int>>& grid, vector<vector<bool>>& visited, int x, int y) {queue<pair<int, int>> que;que.push({x, y});visited[x][y] = true; // 只要加入队列，立刻标记while(!que.empty()) {pair<int ,int> cur = que.front(); que.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;  // 越界了，直接跳过if (!visited[nextx][nexty] && grid[nextx][nexty] == 1) {que.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));for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {cin >> grid[i][j];}}vector<vector<bool>> visited(n, vector<bool>(m, false));int result = 0;for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {if (!visited[i][j] && grid[i][j] == 1) {result++; // 遇到没访问过的陆地，+1bfs(grid, visited, i, j); // 将与其链接的陆地都标记上 true}}}cout << result << endl;
}

题目链接/文章讲解/视频讲解：
https://www.programmercarl.com/kamacoder/0099.%E5%B2%9B%E5%B1%BF%E7%9A%84%E6%95%B0%E9%87%8F%E5%B9%BF%E6%90%9C.html

三、岛屿的最大面积

解题代码C++：

#include <iostream>
#include <vector>
using namespace std;
int count;
int dir[4][2] = {0, 1, 1, 0, -1, 0, 0, -1}; // 四个方向
void dfs(vector<vector<int>>& grid, vector<vector<bool>>& visited, int x, int y) {for (int i = 0; i < 4; i++) {int nextx = x + dir[i][0];int nexty = y + dir[i][1];if (nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size()) continue;  // 越界了，直接跳过if (!visited[nextx][nexty] && grid[nextx][nexty] == 1) { // 没有访问过的 同时 是陆地的visited[nextx][nexty] = true;count++;dfs(grid, visited, nextx, nexty);}}
}int main() {int n, m;cin >> n >> m;vector<vector<int>> grid(n, vector<int>(m, 0));for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {cin >> grid[i][j];}}vector<vector<bool>> visited(n, vector<bool>(m, false));int result = 0;for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {if (!visited[i][j] && grid[i][j] == 1) {count = 1;  // 因为dfs处理下一个节点，所以这里遇到陆地了就先计数，dfs处理接下来的相邻陆地visited[i][j] = true;dfs(grid, visited, i, j); // 将与其链接的陆地都标记上 trueresult = max(result, count);}}}cout << result << endl;}