题解 | 挡住洪水

#include <iostream>
#include <utility>
#include <vector>
using namespace std;


//把bds看作dfs就好了嘿嘿。。。。。
//遇到墙或者越界就返回，否则就上下左右入侵，把0变成墙
void bfs(vector<string>& grid, int x, int y)
{
    int n = grid.size();
    int m = grid[0].size();
    if (x < 0 || x >= n || y < 0 || y >= m || grid[x][y] == '*')
    {
        return;
    }
    grid[x][y] = '*';
    bfs(grid,x-1,y);
    bfs(grid, x, y-1);
    bfs(grid, x+1, y);
    bfs(grid, x, y+1);
}

int main() {
    //错误提示，我忘记cin获取参数了呜呜呜。
    int n,m;
    cin >> n>>m;
    vector<string> grid(n);
    
    
    for (int i = 0; i<n; ++i)
    {
        cin >> grid[i];
    }

  //从四条边界入侵
    for (int i = 0; i<n; ++i)
    {
        if (grid[i][0] == '0')
        {
            bfs(grid, i, 0);
        }
        if (grid[i][m-1] == '0')
        {
            bfs(grid, i, m-1);
        }
    }

    for (int j = 1; j<m-1; ++j)
    {
        if (grid[0][j] == '0')
        {
            bfs(grid, 0, j);
        }
        if (grid[n-1][j] == '0')
        {
            bfs(grid, n-1, j);
        }
    }
    int ans =0;
    for (int i = 1; i<n-1; ++i)
    {
        for (int j = 1; j<m-1; ++j)
        {
            if (grid[i][j] == '0')
            {
                ans++;
            }
        }
    }
    cout << ans;
}
// 64 位输出请用 printf("%lld")