[DFS] Codeforces 510B：Fox And Two Dot

地址：http://codeforces.com/problemset/problem/510/B

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.

Examples

input

3 4AAAAABCAAAAA

output

Yes

input

3 4AAAAABCAAADA

output

No

input

4 4YYYRBYBYBBBYBBBY

output

Yes

input

7 6AAAAABABBBABABAAABABABBBABAAABABBBABAAAAAB

output

Yes

input

2 13ABCDEFGHIJKLMNOPQRSTUVWXYZ

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：给你一张字母图，判断其中存在某种字母成环。

解题思路：DFS，成环，最终又得回到起点，但是有一种不行（从第一个字母@搜到第二个字母@，而第二个字母@又搜到了第一个字母@，不成环，不符合题意）。那么这里我们多加两个参数，来记载当前节点的父节点（上一个字母），当你扫下一个方向时，扫到父节点，那么就跳过，扫下一个方向，如果你找到了你曾经走过的点，那么一定成环了。不回头查找

#include<cstdio>  #include<cstring>  #include<algorithm>  using namespace std;  char a[55][55];  int vis[55][55];  int fx[4]={0,0,1,-1};  int fy[4]={1,-1,0,0};  int n,m,ans;  void dfs(int x,int y,int prex,int prey )//fx，fy为父节点  即上一个查找点 {      if(vis[x][y])      {          ans=1;          return ;    }      vis[x][y]=1;  //每经过一个点 均标记为 1 直到下一个点为 1 即回路    int nx,ny;      for(int i=0;i<4;i++)      {          nx=x+fx[i];          ny=y+fy[i];          if(nx==prex&&ny==prey)//遇到上一个点 因为不能回头 所以继续左右查找            continue;          if(a[nx][ny]==a[x][y])//找到此点  以此点为初始点  继续查找            dfs(nx,ny,x,y);      }  }  int main()  {      while(~scanf("%d %d",&n,&m))      {          ans=0;          memset(vis,0,sizeof(vis));        for(int i=0;i<n;i++)              scanf("%s",a[i]);          for(int i=0;i<n;i++)          {              for(int j=0;j<m;j++)              {                  if(!vis[i][j])//遍历没走过的点                    dfs(i,j,i,j);                  if(ans)  //找到环  跳出                     break;              }              if(ans)  //接着跳出                break;          }     //printf("%s\n",ans?"Yes":"No");//大佬的高级用法        if(ans)              printf("Yes\n");          else              printf("No\n");      }  }