CF 510B(Fox And Two Dots-图上找环)

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.

Sample test(s)

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).

对于每个‘颜色连通块’当成树dfs（不走父边），观察是否有走向已走过点的边

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (50+10)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
int n,m;
char a[MAXN][MAXN];
bool b[MAXN][MAXN]={0};
bool dfs(int i,int j,int t)
{
if (b[i][j]==1) return 1;
b[i][j]=1;
if (a[i][j]==a[i-1][j]&&t!=2) if (dfs(i-1,j,1)) return 1;
if (a[i][j]==a[i+1][j]&&t!=1) if (dfs(i+1,j,2)) return 1;
if (a[i][j]==a[i][j-1]&&t!=4) if (dfs(i,j-1,3)) return 1;
if (a[i][j]==a[i][j+1]&&t!=3) if (dfs(i,j+1,4)) return 1;

return 0;
}
int main()
{
//	freopen("Dots.in","r",stdin);
//	freopen(".out","w",stdout);
MEM(a)
cin>>n>>m;
For(i,n) scanf("%s",a[i]+1);

For(i,n) For(j,m)
{
if (!b[i][j])
{
if (dfs(i,j,0))
{
cout<<"Yes"<<endl;
return 0;
}
}
}
cout<<"No"<<endl;

return 0;
}