Fox And Two Dots --DFS

 Fox And Two Dots

 
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:

These k dots are different: if i ≠ j then di is different from dj.
k is at least 4.
All dots belong to the same color.
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.

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.

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

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).

#include<cstd
4000
io>
#include<cstring>
#include<algorithm>
using namespace std;
int h,w;
int flag;
char map[55][55];
int  pos[55][55];
int ex,ey;
int fx[4]={0,0,1,-1},fy[4]={-1,1,0,0};
void dfs(int x,int y,int step)
{
if(flag)
return ;
pos[x][y]=1;
for(int i=0;i<4;i++)
{
int xx=x+fx[i],yy=y+fy[i];
if(ex==xx&&ey==yy&&step>2)//步数必须大于2否则会出现 BAAB这种类似的情况*/
{
flag=1;
return;
}
if(xx>=0&&yy>=0&&xx<h&&yy<w&&!pos[xx][yy]&&map[x][y]==map[xx][yy])///判断是否越界
dfs(xx,yy,step+1);                                                //颜色是否相同
}
}
int main()
{
scanf("%d%d",&h,&w);
for(int i=0;i<h;i++)
{
scanf("%s",map[i]);
}

flag=0;
for(int i=0;i<h&&!flag;i++)
{
for(int j=0;j<w&&!flag;j++)
{
memset(pos,0,sizeof(pos));
ex=i;ey=j;
dfs(i,j,0);
}
}
if(flag)
printf("Yes");
else
printf("No");

return 0;
}