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.

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

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).

欧拉回路，用bfs搞了一通，发现好乱，学习大神DFS，听房神说还有很简单的方法，明天继续看一下。

 #include <iostream>

 #include<cstdio>

 #include<cstring>

 #include<algorithm>

 #include<queue>

 using namespace std;

 char maze[][];

 int vis[][];

 int n,m;

 int flag = ;

 int to[][] = {{,},{-,},{,},{,-}};

 bool check(int x,int y){

     if(x<||x>=n||y<||y>=m) return false;

     //if(vis[x][y]) return false;

     return true;

 }

 void dfs(int x,int y,int prex,int prey){

     if(!check(x,y)) return;

     vis[x][y] = ;

     int postx,posty;

     for(int i = ; i<; i++){

         postx = x + to[i][];

         posty = y + to[i][];

         if(check(postx,posty)&&maze[postx][posty] == maze[x][y]&&(postx!=prex||posty!=prey)){

             if(vis[postx][posty]){

                 flag = ;

                 return;

             }

             dfs(postx,posty,x,y);

         }

     }

 }

 void input(){

     scanf("%d%d",&n,&m);

     for(int i = ; i<n; i++) scanf("%s",maze[i]);

     for(int i = ; i<n; i++){

         for(int j = ; j<m; j++){

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

                 dfs(i,j,-,-);

             }

         }

     }

     if(flag) printf("Yes\n");

     else printf("No\n");

 }

 int main()

 {

     input();

     return ;

 }