题目传送门
1 /*
2 BFS：三维BFS，加上方向。用dp[x][y][d]记录当前需要的最少转向数
3 */
4 #include <cstdio>
5 #include <algorithm>
6 #include <cstring>
7 #include <queue>
8 #include <cmath>
9 using namespace std;
10
11 const int MAXN = 1e2 + 10;
12 const int INF = 0x3f3f3f3f;
13 struct P
14 {
15 int x, y, z;
16 }now, to;
17 char maze[MAXN][MAXN];
18 int dp[MAXN][MAXN][4];
19 bool inq[MAXN][MAXN][4];
20 int dx[4] = {-1, 1, 0, 0};
21 int dy[4] = {0, 0, -1, 1};
22 int k, x1, y1, x2, y2;
23 int n, m;
24
25 bool check(int x, int y)
26 {
27 if (x >= 1 && x <= n && y >= 1 && y <= m) return true;
28 else return false;
29 }
30
31 bool BFS(void)
32 {
33 memset (dp, 0, sizeof (dp));
34 memset (inq, false, sizeof (inq));
35 queue<P> Q;
36 for (int i=1; i<=n; ++i)
37 {
38 for (int j=1; j<=m; ++j)
39 {
40 for (int l=0; l<4; ++l)
41 {
42 dp[i][j][l] = INF; inq[i][j][l] = false;
43 }
44 }
45 }
46 for (int i=0; i<4; ++i)
47 {
48 dp[x1][y1][i] = 0; inq[x1][y1][i] = true;
49 Q.push ((P) {x1, y1, i});
50 }
51
52 while (!Q.empty ())
53 {
54 now = Q.front (); Q.pop ();
55 for (int i=0; i<4; ++i)
56 {
57 to.x = now.x + dx[i];
58 to.y = now.y + dy[i]; to.z = i;
59 if (check (to.x, to.y) && maze[to.x][to.y] == '.')
60 {
61 int tmp = dp[now.x][now.y][now.z];
62 if (to.z == now.z)
63 {
64 if (tmp < dp[to.x][to.y][to.z])
65 {
66 dp[to.x][to.y][to.z] = tmp;
67 if (!inq[to.x][to.y][to.z])
68 {
69 inq[to.x][to.y][to.z] = true; Q.push (to);
70 }
71 }
72 }
73 else
74 {
75 if (++tmp < dp[to.x][to.y][to.z])
76 {
77 dp[to.x][to.y][to.z] = tmp;
78 if (!inq[to.x][to.y][to.z])
79 {
80 inq[to.x][to.y][to.z] = true; Q.push (to);
81 }
82 }
83 }
84 }
85 }
86 inq[now.x][now.y][now.z] = false;
87 }
88
89 for (int i=0; i<4; ++i)
90 {
91 if (dp[x2][y2][i] <= k) return true;
92 }
93
94 return false;
95 }
96
97 int main(void) //HDOJ 1728 逃离迷宫
98 {
99 // freopen ("HDOJ_1728.in", "r", stdin);
100
101 int t; scanf ("%d", &t);
102 while (t--)
103 {
104 scanf ("%d%d", &n, &m);
105 for (int i=1; i<=n; ++i) scanf ("%s", maze[i] + 1);
106 scanf ("%d%d%d%d%d", &k, &y1, &x1, &y2, &x2);
107 if (BFS ()) puts ("yes");
108 else puts ("no");
109 }
110
111 return 0;
112 }