1 #include<iostream>
 2 #include<cstring>
 3 #include<string>
 4 using namespace std;
 5 
 6 char map[10][10];
 7 int d[][2] = { { 1, 0 }, { -1, 0 }, { 0, 1 }, { 0, -1 } };
 8 int flag,n,m,t,dx,dy;
 9 
10 void dfs(int x,int y,int time)
11 {
12     if (x<1||x>n||y<1||y>m ||flag||time>t)  return;            //出界
13     if (time == t && x==dx && y==dy)
14     {
15         flag = 1;
16         return;
17     }
18     int s1 = x - dx;
19     int s2 = y - dy;
20     int ans = t - time - abs(s1) - abs(s2);       //剪枝，如果当前剩余的所要求的步数减去小狗
21     if (ans<0||ans%2)  return;                    //当前位置与终点的步数不为偶数的话，则结束
22     for (int i = 0; i < 4; i++)
23     {
24         if (map[x + d[i][0]][y + d[i][1]] != ‘X‘)
25         {
26             map[x + d[i][0]][y + d[i][1]] = ‘X‘;
27             dfs(x + d[i][0], y + d[i][1], time+1);
28             map[x + d[i][0]][y + d[i][1]] = ‘.‘;    //这里必须把该点的值还原回来，不然影响后续的dfs
29         }
30     }
31     return;
32 }
33 
34 int main()
35 {
36     int x,y,wall; 
37     while (cin >> n >> m >> t, n && m && t)
38     {
39         if (!m || !n || !t)
40         {
41             cout << "NO" << endl;
42             continue;
43         }
44         flag = 0;
45         wall = 0;
46         for (int i = 1; i <= n;i++)
47         for (int j = 1; j <= m; j++)
48         {
49             cin >> map[i][j];
50             if (map[i][j] == ‘S‘)
51             {
52                 x = i;
53                 y = j;
54             }
55             if (map[i][j] == ‘D‘)
56             {
57                 dx = i;
58                 dy = j;
59             }
60             if (map[i][j] == ‘X‘)
61                 wall++;
62         }
63         if (n*m - wall <t)              //剪枝，如果所有点减去墙小于指定步数，那肯定是不行的
64         {
65             cout << "NO" << endl;
66             continue;
67         }
68         map[x][y] = ‘X‘;
69         dfs(x,y,0);
70         if (flag == 1)  cout << "YES" << endl;
71         else cout << "NO" << endl;
72     }
73     return 0;
74 }