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B. Fox And Two Dots



time limit per test



memory limit per test



input



output


n × m



codeforces-Fox And Two Dots【DFS】（思维）_i++


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.


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



A' form a cycle.

In second sample there is no such cycle.

Y' = Yellow, 'B' = Blue, 'R' = Red).


题意：在n*m的地图上，每个格子涂有不同的颜色（A~Z），问是否有某一种颜色能组成一个回路且长度至少为4. 

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int n,m;
char map[110][110];
bool vis[110][110];
int dx[4]={-1,1,0,0};
int dy[4]={0,0,-1,1};
bool judge(int x,int y)
{
  if(x>=0&&x<n&&y>=0&&y<m)
    return 1;
  return 0;
}
bool DFS(int x,int y,int px,int py,int cnt)
{
  vis[x][y]=1;
  for(int i=0;i<4;i++)
  {
    int nx1=x+dx[i];
    int ny1=y+dy[i];
    if(vis[nx1][ny1]&&judge(nx1,ny1)&&(nx1!=px||ny1!=py)&&cnt>3) // 里面第三个判断条件是保证不会回头计算上一步刚走过的那个点
      return 1;
  }
  char c=map[x][y]; 
  for(int i=0;i<4;i++)
  {
    int nx2=x+dx[i];
    int ny2=y+dy[i];
    if(!vis[nx2][ny2]&&judge(nx2,ny2)&&map[nx2][ny2]==c)
    {
      return DFS(nx2,ny2,x,y,cnt+1); // 这个不是 void 型的 DFS 所以调用自己要加 return 
    }
  }
  return 0;
}
int main()
{
  while(~scanf("%d %d",&n,&m))
  {
    bool flag=0;
    for(int i=0;i<n;i++)
      scanf("%s",map[i]);
    for(int i=0;i<n;i++)
    {
      for(int j=0;j<m;j++)
      {
        memset(vis,0,sizeof(vis));
        if( DFS(i,j,i,j,1) )
        {
          flag=1;
          break;
        }
      }
      if(flag)  break;
    }
    if(flag)
      printf("Yes\n");
    else
      printf("No\n");
  }
  return 0;
}


codeforces-Fox And Two Dots【DFS】（思维）_ide_02