99. 岛屿数量

#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(vector<vector<int>>& grid, int x, int y) {// cout<<"x："<<x<<"y："<<y<<":"<<grid[x][y]<<endl;grid[x][y] = 0; 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() && grid[nextx][nexty] == 1){ dfs(grid, 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];}}int result = 0;for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {if (grid[i][j] == 1) {result++; // 遇到没访问过的陆地，+1dfs(grid, i, j); // 将与其链接的陆地标记0，避免重复计数}}}cout << result << endl;
}

深搜，首次访问到时计数，并将相邻元素清零，避免重复统计。

#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(vector<vector<int>>& grid, int x, int y) {queue<pair<int, int>> que;que.push({x, y});grid[x][y] = 0; while(!que.empty()) {auto [curx,cury] = que.front(); que.pop();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 (grid[nextx][nexty] == 1) {que.push({nextx, nexty});grid[nextx][nexty] = 0; // 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,  i, j); // 将与其链接的陆地都标记上 true}}}cout << result << endl;
}

广搜，每次向队列中push一圈元素，不断向外膨胀。

100. 岛屿的最大面积

// 版本二
#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, int x, int y) {grid[x][y] = 0; // 标记访问过count++;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() && grid[nextx][nexty] == 1){dfs(grid, 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];}}int result = 0;for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {if (grid[i][j] == 1) {count = 0; // 因为dfs处理当前节点，所以遇到陆地计数为0，进dfs之后在开始从1计数dfs(grid, i, j); // 将与其链接的陆地都标记上 trueresult = max(result, count);}}}cout << result << endl;
}

深搜，用全局变量累计单个岛屿面积。主程序里用result选出最大。