B. Fox And Two Dots

题记：本题目为cf的B题，一个图中是否有相同字母围城的环，思路为DFS,深度优先遍历，只要一个遍历的再次回到已经遍历的地方，那么久一定能组成环，具体见代码实现。

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, ..., d_k a cycle if and only if it meets the following condition:

1. These k dots are different: if i ≠ j then d_i is different from d_j.
2. k is at least 4.
3. All dots belong to the same color.
4. For all 1 ≤ i ≤ k - 1: d_i and d_i + 1 are adjacent. Also, d_k 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
代码：
    #include <cstdio>
    #include <cstring>
    #include <iostream>
    #include <algorithm>
    using namespace std;
    int t[4][2]={{1,0},{-1,0},{0,-1},{0,1}};
    int vis[55][55];
    int mark;
    int m,n;
    char s[55][55];
    void dfs(int x1,int y1,int x2,int y2)
    {
        vis[x1][y1]=1;
        for(int i=0;i<4;i++)
        {
            int X=x1+t[i][0];
            int Y=y1+t[i][1];
            if(X>=0&&Y>=0&&X<n&&Y<m&&(X!=x2||Y!=y2)&&s[x1][y1]==s[X][Y])
            {
                if(vis[X][Y]==1)
               {

               mark=1;break;
               }
               dfs(X,Y,x1,y1);
            }
        }
    }
    int main()
    {
        scanf("%d%d",&n,&m);
            memset(vis,0,sizeof(vis));
            memset(s,0,sizeof(s));
            for(int i = 0; i < n ; i++)
            {
                scanf("%s",s[i]);
            }
            mark = 0;
            for(int i = 0; i < n; i++)
            {
                for(int j = 0; j < m; j++)
                {
                    if(!vis[i][j])
                    {
                        dfs(i,j,-1,-1);
                        if(mark)
                        {
                            printf("Yes\n");
                            return 0;
                        }
                    }
                }
            }
            printf("No\n");
        return 0;
    }