B. Fox And Two Dots( Codeforces Round #290 (Div. 2))

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).

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>

using namespace std;

int n,m;
char map[51][51];
int v[51][51];
int jx[] = {0,1,0,-1};
int jy[] = {1,0,-1,0};
int flag = 0;
struct node
{
    int x;
    int y;
};

void DFS(int xx,int yy,int x1,int y1)
{

    for(int i=0; i<4; i++)
    {
        if(flag == 1)
        {
            return ;
        }
        int x,y;
        x = xx + jx[i];
        y = yy + jy[i];
        if(v[x][y] == 1 && map[xx][yy] == map[x][y] && x1!=x && y1!=y && x>=0 && x<n && y>=0 && y<m)
        {
            printf("Yes\n");
            flag = 1;
            return ;
        }
        if(v[x][y] == 0)
        {
            if(map[xx][yy] == map[x][y] && x>=0 && x<n && y>=0 && y<m)
            {
                //printf("x = %d   y = %d\n",x,y);
                v[x][y] = 1;
                DFS(x,y,xx,yy);
            }

        }
    }
}

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        flag = 0;
        memset(v,0,sizeof(v));
        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++)
            {
                if(v[i][j] == 0 && flag == 0)
                {
                    DFS(i,j,i,j);
                }
            }
        }
        if(flag == 0)
        {
            printf("No\n");
        }

    }
    return 0;
}