codeforce-B. Fox And Two Dots

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:

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).
一题我用的是dfs 其实仔细想想，你会发现从一个点跑，你可以想象一下其实dfs就像一只会分身

的老鼠，到达一个点只要四周可以跑，就会分身，一直跑下去，只要存在一个回路，它就会发现已经走过

然后就会汇报有回路，如果跑到最后都没有回路，说明从这个点和所有跑过的路都不会有回路，因为只要

存在回路，它就会发现其他老鼠跑过的路 值得注意的时不要走回路 

<span style="font-family:SimSun;">#include<cstdio>
#include<cstring>
char s[100][100];
int vis[100][100];//用于标记
int n,m;//n行m列
int flag;
int dx[4]={1,-1,0,0};
int dy[4]={0,0,1,-1};
int judge(int a,int b)
{
if(a>=0&&a<n&&b>=0&&b<m)
return 1;
return 0;
}
void dfs(int x,int y,char o,int xx,int yy)//记录上一步走的路
{
int i;
for(i=0;i<4;++i)
{
int ex=x+dx[i];
int ey=y+dy[i];
if(judge(ex,ey)&&s[ex][ey]==o)
{
if(ex==xx&&ey==yy)
continue;
if(vis[ex][ey])
{
flag=1;
return ;
}
vis[ex][ey]=1;
dfs(ex,ey,o,x,y);
}
}
}
int main()
{
scanf("%d%d",&n,&m);
int i,j;
for(i=0;i<n;++i)
scanf("%s",s[i]);//这样就很容易形成地图
memset(vis,0,sizeof(vis));
for(i=0;i<n;++i)
for(j=0;j<m;++j)
{
if(!vis[i][j])
{
flag=0;
vis[i][j]=1;
dfs(i,j,s[i][j],-1,-1);
if(flag)
{
printf("Yes\n");
return 0;
}
}
}
printf("No\n");
return 0;
} </span>