1 /*
2 BFS+模拟：dp[i][j][p] 表示走到i，j，方向为p的步数为多少；
3 BFS分4种情况入队，最后在终点4个方向寻找最小值:)
4 */
5 #include <cstdio>
6 #include <iostream>
7 #include <algorithm>
8 #include <cstring>
9 #include <string>
10 #include <queue>
11 using namespace std;
12
13 const int MAXN = 1e2 + 10;
14 const int INF = 0x3f3f3f3f;
15 int dir[4][4][2] = {
16 {{0, -1}, {0, 1}, {-1, 0}, {1, 0}},
17 {{1, 0}, {0, -1}, {0, 1}, {-1, 0}},
18 {{-1, 0}, {1, 0}, {0, -1}, {0, 1}},
19 {{0, 1}, {-1, 0}, {1, 0}, {0, -1}}
20 };
21 int dp[12][12][4];
22 int dp2[12][12][4];
23 int ans[12][12][4];
24 char maze[12][12];
25 int sx, sy, ex, ey;
26 int n, m, P;
27
28 bool ok(int nx, int ny, int np)
29 {
30 if (nx < 0 || nx >= n || ny < 0 || ny >= m) return false;
31 if (maze[nx][ny] == '*' || dp[nx][ny][np] != -1) return false;
32
33 return true;
34 }
35
36 void BFS(void)
37 {
38 int x, y, nx, ny, p, np;
39 queue<int> q;
40 q.push (sx); q.push (sy); q.push (0);
41
42 while (!q.empty ())
43 {
44 x = q.front (); q.pop ();
45 y = q.front (); q.pop ();
46 p = q.front (); q.pop ();
47
48 if (x == ex && y == ey && ans[x][y][p] == -1)
49 ans[x][y][p] = dp[x][y][p];
50
51 nx = x + dir[dp[x][y][p]/P%4][p][0];
52 ny = y + dir[dp[x][y][p]/P%4][p][1];
53 if (ok (nx, ny, p)) //move robot, not move dir
54 {
55 dp[nx][ny][p] = dp[x][y][p] + 1;
56 q.push (nx); q.push (ny); q.push (p);
57 }
58
59 np = p + 1;
60 if (np > 3) np = 0;
61 if (ok (x, y, np)) //not move robot, move dir to right
62 {
63 dp[x][y][np] = dp[x][y][p] + 1;
64 q.push (x); q.push (y); q.push (np);
65 }
66
67 np = p - 1;
68 if (np < 0) np = 3;
69 if (ok (x, y, np)) //not move robot, move dir to left
70 {
71 dp[x][y][np] = dp[x][y][p] + 1;
72 q.push (x); q.push (y); q.push (np);
73 }
74
75 if (dp[x][y][p] <= 200) //not move
76 {
77 dp[x][y][p]++;
78 q.push (x); q.push (y); q.push (p);
79 }
80 }
81
82 return ;
83 }
84
85 int main(void) //ZOJ 3865 Superbot
86 {
87 //freopen ("F.in", "r", stdin);
88
89 int t;
90 scanf ("%d", &t);
91 while (t--)
92 {
93 memset (dp, -1, sizeof (dp));
94 memset (ans, -1, sizeof (ans));
95
96 scanf ("%d%d%d", &n, &m, &P);
97
98 for (int i=0; i<n; ++i) scanf ("%s", &maze[i]);
99 for (int i=0; i<n; ++i)
100 {
101 for (int j=0; j<m; ++j)
102 {
103 if (maze[i][j] == '@') sx = i, sy = j;
104 else if (maze[i][j] == '$') ex = i, ey = j;
105 }
106 }
107
108 dp[sx][sy][0] = 0;
109 BFS ();
110
111 int mn = INF;
112 for (int i=0; i<4; ++i)
113 {
114 if (ans[ex][ey][i] == -1) continue;
115 mn = min (mn, ans[ex][ey][i]);
116 }
117
118 if (mn == INF) printf ("YouBadbad
");
119 else printf ("%d
", mn);
120 }
121
122 return 0;
123 }