CF B. Fox And Two Dots

B. Fox And Two Dots

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Fox Ciel is playing a mobile puzzle game called "Two Dots". The basic levels are played on a board of size n × m cells, like this:

Each cell contains a dot that has some color. We will use different uppercase Latin characters to express different colors.

The key of this game is to find a cycle that contain dots of same color. Consider 4 blue dots on the picture forming a circle as an example. Formally, we call a sequence of dots d1, d2, ..., dk a cycle if and only if it meets the following condition:

1. These k dots are different: if i ≠ j then di is different from dj.
2. k is at least 4.
3. All dots belong to the same color.
4. For all 1 ≤ i ≤ k - 1: di and di + 1 are adjacent. Also, dk and d1 should also be adjacent. Cells x and y are called adjacent if they share an edge.

Determine if there exists a cycle on the field.

Input

The first line contains two integers n and m (2 ≤ n, m ≤ 50): the number of rows and columns of the board.

Then n lines follow, each line contains a string consisting of m characters, expressing colors of dots in each line. Each character is an uppercase Latin letter.

Output

Output "Yes" if there exists a cycle, and "No" otherwise.

Sample test(s)

input

3 4
AAAA
ABCA
AAAA

output

Yes

input

3 4
AAAA
ABCA
AADA

output

No

input

4 4
YYYR
BYBY
BBBY
BBBY

output

Yes

input

7 6
AAAAAB
ABBBAB
ABAAAB
ABABBB
ABAAAB
ABBBAB
AAAAAB

output

Yes

input

2 13
ABCDEFGHIJKLM
NOPQRSTUVWXYZ

output

No

Note

In first sample test all 'A' form a cycle.

In second sample there is no such cycle.

The third sample is displayed on the picture above ('Y' = Yellow, 'B' = Blue, 'R' = Red).

 题意：判断图中是否存在长度大于4的回路(由同一种字母组成的)。

思路：DFS，注意每次不往回搜索，直到搜索结束或者搜到有标记的位置停止。

 1 #include<iostream>

 2 #include<cstdio>

 3 #include<cstdlib>

 4 #include<cstring>

 5 #include<string>

 6 #include<queue>

 7 #include<algorithm>

 8 #include<map>

 9 #include<iomanip>

10 #include<climits>

11 #include<string.h>

12 #include<numeric>

13 #include<cmath>

14 #include<stdlib.h>

15 #include<vector>

16 #include<stack>

17 #include<set>

18 #define FOR(x, b, e)  for(int x=b;x<=(e);x++)

19 #define REP(x, n)     for(int x=0;x<(n);x++)

20 #define INF 1e7

21 #define MAXN 100010

22 #define maxn 1000010

23 #define Mod 1000007

24 #define N 1010

25 using namespace std;

26 typedef long long LL;

27 

28 

29 char G[55][55];

30 int vis[55][55];

31 int dx[] = {-1, 0, 0, 1}, dy[] = {0, 1, -1, 0};

32 int n, m;

33 bool ok;

34 

35 void dfs(int x,int y, int d) 

36 {

37     vis[x][y] = 1;

38     REP(i, 4) {

39         if (i == 3 - d) continue;

40         int nx = x + dx[i];

41         int ny = y + dy[i];

42         if (nx == 0 || ny == 0 || ny == m+1 || nx == n+1) continue;

43         if (G[x][y] == G[nx][ny]) {

44             if (vis[nx][ny]) {

45                 ok = true;

46                 return;

47             }

48             else dfs(nx, ny, i);

49         }

50     }

51 }

52 

53 int main()

54 {

55     cin >> n >> m;

56     FOR(i, 1, n)

57         cin >> G[i] + 1;

58     FOR(i, 1, n)

59         FOR(j, 1, m) {

60             if (!vis[i][j]) 

61                 dfs(i, j, 4);

62         }

63     puts(ok ? "Yes" : "No");

64     return 0;

65 }