CodeForces - 510B 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.

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 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).

ps：直接dfs，只要走到已经走过的点就说明已经成环

注意：有可能出现从这个点走出去一步又回到这个点的情况，所以需要标记一下当前点的下一步坐标不能和当前点的上一步坐标相同

代码：

#include<stdio.h>
#include<string.h>

char s[55][55];
int vis[55][55];
int n,m,flag;
int to[][2]= {0,1,0,-1,1,0,-1,0};
char k;

void dfs(int x,int y,int pre_x,int pre_y)//pre_x为当前点的上一步点的x轴，pre_y为当前点的上一步点的y轴
{
if(flag)
return ;
for(int i=0; i<4; i++)
{
int xx=x+to[i][0],yy=y+to[i][1];
if(s[xx][yy]==k&&xx>=0&&xx<n&&yy>=0&&yy<m)
{
if(xx==pre_x&&yy==pre_y)
continue;
else if(vis[xx][yy])
{
flag=1;
return ;
}
else if(!vis[xx][yy])
{
vis[xx][yy]=1;
dfs(xx,yy,x,y);
}
}
}
}

int main()
{
flag=0;
scanf("%d%d",&n,&m);
for(int i=0; i<n; i++)
scanf("%s",s[i]);
for(int i=0; i<n; i++)
{
for(int j=0; j<m; j++)
{
memset(vis,0,sizeof(vis));
k=s[i][j],vis[i][j]=1;
for(int t=0; t<4; t++)
{
int xx=i+to[t][0],yy=j+to[t][1];
if(s[xx][yy]==k&&xx>=0&&xx<n&&yy>=0&&yy<m)
dfs(xx,yy,i,j);
}
if(flag)
break;
}
if(flag)
break;
}
if(flag)
printf("Yes\n");
else
printf("No\n");
return 0;
}