E

题目链接：http://acm.hust.edu.cn/vjudge/contest/127946#problem/E

分析：存在某个字符能围成封闭图形，则说明Yes，否则是No。

Sample Input
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

Hint
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).

*********************************************

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

#define N 1200

int v[N][N],n,m,ww,cnt;
char s[60][60];
int dir[4][2]= { {1,0},{-1,0},{0,1},{0,-1} };

void q(char ch,int x,int y,int X,int Y)
{
    int i,xx,yy;
    if(ww==1)
        return ;

    for(i=0; i<4; i++)
    {
        xx=x+dir[i][0];
        yy=y+dir[i][1];
        if(xx==X&&yy==Y)
            continue ;
        if(xx>=0&&xx<n&&yy>=0&&yy<m&&s[xx][yy]==ch&&v[xx][yy])
        {
            ww=1;
           return ;
        }
         if(xx>=0&&xx<n&&yy>=0&&yy<m&&s[xx][yy]==ch&&!v[xx][yy])
        {
            v[xx][yy]=1;
            q(ch,xx,yy,x,y);
        }
    }
}

int main()
{
    int i,j;

    while(scanf("%d %d", &n, &m) != EOF)
    {
        ww=0,cnt=0;
        memset(v,0,sizeof(v));
        for(i=0; i<n; i++)
            scanf("%s", s[i]);

        for(i=0; i<n; i++)
        {
            for(j=0; j<m; j++)
            {
                if(v[i][j]==0)
                {
                    v[i][j]=1;
                    q(s[i][j],i,j,0,0);
                    if(ww==1)
                    {
                       cnt=1;
                       break;
                    }
                }
            }
        }
        if(cnt==1)
            printf("Yes
");
        else
            printf("No
");
    }
    return 0;
}