CF 510B(Fox And Two Dots-图上找环)

B. Fox And Two Dots

time limit per test

memory limit per test

input

output

n × m

Each cell contains a dot that has some color. We will use different uppercase Latin characters to express different colors.

d1, d2, ..., dk a cycle

1. kdots are different: ifi≠jthendiis different fromdj.
2. k
3. All dots belong to the same color.
4. 1 ≤i≤k- 1:dianddi+ 1 are adjacent. Also,dkandd1 should also be adjacent. Cellsxandy

cycle

Input

n and m (2 ≤ n, m ≤ 50): the number of rows and columns of the board.

n lines follow, each line contains a string consisting of m

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

A' form a cycle.

In second sample there is no such cycle.

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;
}