1 /*
2 题意：告诉起点终点，踩一次, '.'变成'X'，再踩一次，冰块破碎，问是否能使终点冰破碎
3 DFS：如题解所说，分三种情况：1. 如果两点重合，只要往外走一步再走回来就行了；2. 若两点相邻，
4 那么要不就是踩一脚就破了或者踩一脚走开再走回来踩一脚破了；3. 普通的搜索看是否能到达，
5 若能还是要讨论终点踩几脚的问题:)
6 DFS 耗时大，险些超时，可以用BFS来做
7 */
8 #include <cstdio>
9 #include <algorithm>
10 #include <cstring>
11 #include <string>
12 #include <iostream>
13 using namespace std;
14
15 const int MAXN = 5e2 + 10;
16 const int INF = 0x3f3f3f3f;
17 int n, m;
18 int sx, sy, ex, ey;
19 char maze[MAXN][MAXN];
20 bool vis[MAXN][MAXN];
21 int dx[4] = {1, -1, 0, 0};
22 int dy[4] = {0, 0, -1, 1};
23
24 void DFS(int x, int y)
25 {
26 vis[x][y] = true;
27 if (maze[x][y] == 'X') return ;
28
29 for (int i=0; i<4; ++i)
30 {
31 int tx = x + dx[i];
32 int ty = y + dy[i];
33
34 if (tx <= n && tx >= 1 && ty <= m && ty >= 1 && !vis[tx][ty])
35 {
36 DFS (tx, ty);
37 }
38 }
39
40 return ;
41 }
42
43 int main(void) //Codeforces Round #301 (Div. 2) C. Ice Cave
44 {
45 //freopen ("C.in", "r", stdin);
46
47 while (scanf ("%d%d", &n, &m) == 2)
48 {
49 memset (vis, 0, sizeof (vis));
50 for (int i=1; i<=n; ++i)
51 {
52 scanf ("%s", maze[i]+1);
53 }
54 scanf ("%d%d", &sx, &sy);
55 scanf ("%d%d", &ex, &ey);
56
57 int cnt = 0; bool flag = false;
58 for (int i=0; i<4; ++i)
59 {
60 int tx = ex + dx[i];
61 int ty = ey + dy[i];
62 if (tx == sx && ty == sy) flag = true;
63 if (tx <= n && tx >= 1 && ty <= m && ty >= 1 && maze[tx][ty] == '.') cnt++;
64 }
65
66 if (sx == ex && sy == ey)
67 {
68 if (cnt >= 1) puts ("YES");
69 else puts ("NO");
70 }
71 else if (flag)
72 {
73 if (maze[ex][ey] == 'X') puts ("YES");
74 else
75 {
76 if (cnt >= 1) puts ("YES");
77 else puts ("NO");
78 }
79 }
80 else
81 {
82 maze[sx][sy] = '.';
83 DFS (sx, sy);
84 if (vis[ex][ey])
85 {
86 if (maze[ex][ey] == 'X') puts ("YES");
87 else if (cnt >= 2) puts ("YES");
88 else puts ("NO");
89 }
90 else puts ("NO");
91 }
92 }
93
94 return 0;
95 }
1 #include <cstdio> 2 #include <algorithm> 3 #include <cstring> 4 #include <queue> 5 #include <string> 6 #include <iostream> 7 using namespace std; 8 9 const int MAXN = 5e2 + 10; 10 const int INF = 0x3f3f3f3f; 11 int n, m; 12 int sx, sy, ex, ey; 13 char maze[MAXN][MAXN]; 14 bool vis[MAXN][MAXN]; 15 int dx[4] = {1, -1, 0, 0}; 16 int dy[4] = {0, 0, -1, 1}; 17 18 bool BFS(void) 19 { 20 queue<pair<int, int> > Q; 21 Q.push (make_pair (sx, sy)); 22 23 while (!Q.empty ()) 24 { 25 int x = Q.front ().first; int y = Q.front ().second; Q.pop (); 26 for (int i=0; i<4; ++i) 27 { 28 int tx = x + dx[i]; 29 int ty = y + dy[i]; 30 31 if (tx == ex && ty == ey) return true; 32 33 if (tx <= n && tx >= 1 && ty <= m && ty >= 1 && maze[tx][ty] == '.') 34 { 35 maze[tx][ty] = 'X'; 36 Q.push (make_pair (tx, ty)); 37 } 38 } 39 } 40 41 return false; 42 } 43 44 int main(void) //Codeforces Round #301 (Div. 2) C. Ice Cave 45 { 46 //freopen ("C.in", "r", stdin); 47 48 while (scanf ("%d%d", &n, &m) == 2) 49 { 50 memset (vis, 0, sizeof (vis)); 51 for (int i=1; i<=n; ++i) 52 { 53 scanf ("%s", maze[i]+1); 54 } 55 scanf ("%d%d", &sx, &sy); 56 scanf ("%d%d", &ex, &ey); 57 58 int cnt = 0; bool flag = false; 59 for (int i=0; i<4; ++i) 60 { 61 int tx = ex + dx[i]; 62 int ty = ey + dy[i]; 63 if (tx == sx && ty == sy) flag = true; 64 if (tx <= n && tx >= 1 && ty <= m && ty >= 1 && maze[tx][ty] == '.') cnt++; 65 } 66 67 if (sx == ex && sy == ey) 68 { 69 if (cnt >= 1) puts ("YES"); 70 else puts ("NO"); 71 } 72 else if (flag) 73 { 74 if (maze[ex][ey] == 'X') puts ("YES"); 75 else 76 { 77 if (cnt >= 1) puts ("YES"); 78 else puts ("NO"); 79 } 80 } 81 else 82 { 83 maze[sx][sy] = '.'; 84 //DFS (sx, sy); 85 if (BFS () == true) 86 { 87 if (maze[ex][ey] == 'X') puts ("YES"); 88 else if (cnt >= 2) puts ("YES"); 89 else puts ("NO"); 90 } 91 else puts ("NO"); 92 } 93 } 94 95 return 0; 96 }