CF290-B

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, ..., d_k a cycle if and only if it meets the following condition:

1. These k dots are different: if i ≠ j then d_i is different from d_j.
2. k is at least 4.
3. All dots belong to the same color.
4. For all 1 ≤ i ≤ k - 1: d_i and d_i + 1 are adjacent. Also, d_k 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).

给定n*m的方格,每个小格子里有A~Z的字母,代表一种颜色,问这个n*m的方格里是否包含一个颜色相同的环.

枚举方格中的每一个格子以其作为起点进行深度优先遍历,看是否能回到起点,能的话就能找到这么一个环,否则就不行.

判断是否回到起点的一个小技巧是:先记录下起点位置,深搜的过程将路径给'堵'住,如果到达一个"新位置"如果可以到达的

下一位置是起点,并且颜色相同.那么这个环存在.这样做的目的是为了防止递归回溯的时候会"误判"下一位置是起点从而认为存在环.

需要注意一点是路径长度要大于3,故还需要记录长度

#include <iostream>

using namespace std;

int count;

int n,m;

bool flag=false;

int dir1[4]={1,-1,0,0};

int dir2[4]={0,0,1,-1};

int use[100][100];

char map[100][100];

int x,y;

int begin,end;

#include <string.h>

bool jud(int i,int j)

{

    if(i<=0||i>n||j<=0||j>m)

        return false;

    return true;

}

void dfs(int a,int b,int k)

{

    if(flag)

        return;

    for(int i=0;i<4;i++)

    {

        if(a+dir1[i]==begin&&b+dir2[i]==end&&use[a][b]==0&&(int)map[a+dir1[i]][b+dir2[i]]==k&&count>=3)

        {

            flag=true;

            break;

        }

        if((int)map[a+dir1[i]][b+dir2[i]]==k&&use[a+dir1[i]][b+dir2[i]]==0&&jud(a+dir1[i],b+dir2[i]))

        {

            use[a][b]=1;

            count++;

            dfs(a+dir1[i],b+dir2[i],k);

            count--;

        }

    }

}



int main()

{

    cin>>n>>m;

    x=y=0;



    memset(map,'.',sizeof(map));

    for(int i=1;i<=n;i++)

    {

        for(int j=1;j<=m;j++)

        {

           cin>>map[i][j];

        }

    }

    for(int i=1;i<=n;i++)

    {

        for(int j=1;j<=m;j++)

        {

            memset(use,0,sizeof(use));

            begin=i;

            end=j;

            count=0;

            dfs(i,j,(int)map[i][j]);

        }

    }

    if(flag)

        cout<<"Yes"<<endl;

    else

        cout<<"No"<<endl;

    return 0;

}