就是在输入的时候把 ‘o’ 的放在队里，然后，直接BFS就可以了。感觉是水题。

#include
     
      #include
      
       using namespace std;#define pa pair
       
        const int maxn = 1e3 + 10;char cc[maxn][maxn];int xx[] = { 0, 1, -1, 0, 0 };int yy[] = { 0, 0, 0, 1, -1 };char kk[] = " udlr";int n, m, ans;queue
        
         q;int main(){    cin >> n >> m;    for (int i = 1; i <= n; ++i){        for (int j = 1; j <= m; ++j)        {            cin >> cc[i][j];            if (cc[i][j] == 'o'){                 pair
         
          ss;                ss.first = i;    ss.second = j;                q.push(ss);            }        }    }    while (q.size()){        pair
          
           ss = q.front(); q.pop(); ans++; for (int i = 1; i <= 4; ++i){ int x = ss.first + xx[i], y = ss.second + yy[i]; if ((x >= 1 && x <= n&&y >= 1 && y <= m) && cc[x][y] == kk[i]){ pair
           
            h; h.first = x; h.second = y; q.push(h); } } } cout << ans << endl;}