给定一幅地图，其中有水域，有陆地。被水域完全环绕的陆地是岛屿。有些岛屿上埋藏有宝藏，这些有宝藏的点也被标记出来了。本题就请你统计一下，给定的地图上一共有多少岛屿，其中有多少是有宝藏的岛屿。

输入格式：

输入第一行给出 2 个正整数 N 和 M（1<N×M≤105），是地图的尺寸，表示地图由 N 行 M 列格子构成。随后 N 行，每行给出 M 位个位数，其中 0 表示水域，1 表示陆地，2-9 表示宝藏。
注意：两个格子共享一条边时，才是“相邻”的。宝藏都埋在陆地上。默认地图外围全是水域。

输出格式：

在一行中输出 2 个整数，分别是岛屿的总数量和有宝藏的岛屿的数量。

输入样例：

10 11
01000000151
11000000111
00110000811
00110100010
00000000000
00000111000
00114111000
00110010000
00019000010
00120000001

输出样例：

7 2

 

#include <bits/stdc++.h>
using namespace std;const int dx[] = {-1, 1, 0, 0};
const int dy[] = {0, 0, -1, 1};void bfs(int x, int y, vector<vector<char>>& grid, bool& hasTreasure) {int n = grid.size();int m = grid[0].size();queue<pair<int, int>> q;q.push({x, y});// 标记当前格子为已访问grid[x][y] = '0';while (!q.empty()) {auto [cx, cy] = q.front();q.pop();// 检查当前格子是否是宝藏if (grid[cx][cy] >= '2' && grid[cx][cy] <= '9') {hasTreasure = true;}// 遍历四个方向for (int i = 0; i < 4; i++) {int nx = cx + dx[i];int ny = cy + dy[i];if (nx >= 0 && nx < n && ny >= 0 && ny < m && grid[nx][ny] != '0') {grid[nx][ny] = '0'; // 标记为已访问q.push({nx, ny});}}}
}void solve() {int n, m;cin >> n >> m;vector<vector<char>> grid(n, vector<char>(m));// 读取地图for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {cin >> grid[i][j];}}int totalIslands = 0;int treasureIslands = 0;// 遍历地图for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {if (grid[i][j] != '0') {bool hasTreasure = false;bfs(i, j, grid, hasTreasure);totalIslands++;if (hasTreasure) {treasureIslands++;}}}}// 输出结果cout << totalIslands << " " << treasureIslands << endl;
}int main() {ios::sync_with_stdio(0);cin.tie(0), cout.tie(0);solve();return 0;
}

#include <bits/stdc++.h>
using namespace std;
const int dx[] = {-1, 1, 0, 0};
const int dy[] = {0, 0, -1, 1};
void dfs(int x, int y, vector<vector<char>>& grid, bool& hasTreasure) {int n = grid.size();int m = grid[0].size();if (grid[x][y] >= '2' && grid[x][y] <= '9') {hasTreasure = true;}grid[x][y] = '0';for (int i = 0; i < 4; i++) {int nx = x + dx[i];int ny = y + dy[i];if (nx >= 0 && nx < n && ny >= 0 && ny < m && grid[nx][ny] != '0') {dfs(nx, ny, grid, hasTreasure);}}
}
void solve() {int n, m;cin >> n >> m;vector<vector<char>> grid(n, vector<char>(m));for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {cin >> grid[i][j];}}int totalIslands = 0;int treasureIslands = 0;for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {if (grid[i][j] != '0') {bool hasTreasure = false;dfs(i, j, grid, hasTreasure);totalIslands++;if (hasTreasure) {treasureIslands++;}}}}cout << totalIslands << " " << treasureIslands << endl;
}int main() {ios::sync_with_stdio(0);cin.tie(0), cout.tie(0);solve();return 0;
}