题目来源：104. 建造最大岛屿
C++题解：先用深度优化算法计算每个孤岛的面积，然后给每个孤岛编号（如果孤岛是U型的，防止面积加重复了），再去遍历每个海水区域，计算最大面积。
#include <iostream>
#include <vector>
#include <map>
#include <set>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, vector<vector<int>> &vis, int &id, int N, int M, int i, int j){if(vis[i][j] > 0 || grid[i][j] == 0) return;vis[i][j] = id;for(int k = 0; k < 4; k++){int tmpi = i + dir[k][0];int tmpj = j + dir[k][1];if(tmpi >= 0 && tmpi < N && tmpj >= 0 && tmpj < M){dfs(grid, vis, id, N, M, tmpi, tmpj);}}
}int main(){
//输入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 id = 1;vector<vector<int>> vis(N, vector<int>(M, 0));// 存id，标记访问过for(int i = 0; i < N; i++) {for(int j = 0; j < M; j++){if(vis[i][j] > 0 || grid[i][j] == 0) continue;dfs(grid, vis, id, N, M, i, j);id++;}}map<int,int> squ; //根据相同的id，计算面积int res = 0;for(int i = 0; i < N; i++) {for(int j = 0; j < M; j++){if(grid[i][j] == 0) continue;squ[vis[i][j]]++;// 找最大面积的孤岛，因为有可能孤岛比连着的面积大res = max(squ[vis[i][j]], res);}}for(int i = 0; i < N; i++) {for(int j = 0; j < M; j++){if(vis[i][j] > 0) continue;int tmpsqu = 1;set<int> dirid;//遍历一下周围的idfor(int k = 0; k < 4; k++) {int tmpi = i + dir[k][0];int tmpj = j + dir[k][1];if(tmpi >= 0 && tmpi < N && tmpj >= 0 && tmpj < M){dirid.insert(vis[tmpi][tmpj]);}}// 不同id的面积相加for(auto it:dirid){tmpsqu = tmpsqu + squ[it];}res = max(res, tmpsqu);}}cout<<res;return 0;
}大概思路一致，代码来源代码随想录
#include <iostream>
#include <vector>
#include <unordered_set>
#include <unordered_map>
using namespace std;
int n, m;
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, int mark) {if (visited[x][y] || grid[x][y] == 0) return; // 终止条件：访问过的节点 或者 遇到海水visited[x][y] = true; // 标记访问过grid[x][y] = mark; // 给陆地标记新标签count++;for (int i = 0; i < 4; i++) {int nextx = x + dir[i][0];int nexty = y + dir[i][1];if (nextx < 0 || nextx >= n || nexty < 0 || nexty >= m) continue; // 越界了，直接跳过dfs(grid, visited, nextx, nexty, mark);}
}int main() {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)); // 标记访问过的点unordered_map<int ,int> gridNum;int mark = 2; // 记录每个岛屿的编号bool isAllGrid = true; // 标记是否整个地图都是陆地for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {if (grid[i][j] == 0) isAllGrid = false;if (!visited[i][j] && grid[i][j] == 1) {count = 0;dfs(grid, visited, i, j, mark); // 将与其链接的陆地都标记上 truegridNum[mark] = count; // 记录每一个岛屿的面积mark++; // 记录下一个岛屿编号}}}if (isAllGrid) {cout << n * m << endl; // 如果都是陆地，返回全面积return 0; // 结束程序}// 以下逻辑是根据添加陆地的位置，计算周边岛屿面积之和int result = 0; // 记录最后结果unordered_set<int> visitedGrid; // 标记访问过的岛屿for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {count = 1; // 记录连接之后的岛屿数量visitedGrid.clear(); // 每次使用时，清空if (grid[i][j] == 0) {for (int k = 0; k < 4; k++) {int neari = i + dir[k][1]; // 计算相邻坐标int nearj = j + dir[k][0];if (neari < 0 || neari >= n || nearj < 0 || nearj >= m) continue;if (visitedGrid.count(grid[neari][nearj])) continue; // 添加过的岛屿不要重复添加// 把相邻四面的岛屿数量加起来count += gridNum[grid[neari][nearj]];visitedGrid.insert(grid[neari][nearj]); // 标记该岛屿已经添加过}}result = max(result, count);}}cout << result << endl;}
