【codeforce-510B】fox and two dots

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.

Example
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<stdio.h>#include<string.h>#include<algorithm>using namespace std;int n,m,flag=0;char c[55][55];int vis[55][55];int s;int a,b;int fx[4]={0,0,1,-1},fy[4]={1,-1,0,0};void dfs(int x,int y,int nl){    if(s) return ;    vis[x][y]=1;    for(int i=0;i<4;i++)    {        int xx=x+fx[i],yy=y+fy[i];        if(xx==a&&yy==b&&nl>=3)        {            s=1;           return;        }        if(xx>=0&&yy>=0&&xx<n&&yy<m&&!vis[xx][yy]&&c[xx][yy]==c[x][y])        {            dfs(xx,yy,nl+1);        }    }}int main(){    scanf("%d%d",&n,&m);    memset(vis,0,sizeof(vis));    for(int i=0;i<n;i++)    scanf("%s",c[i]);    s=0;    for(int j=0;j<n;j++)    {        for(int i=0;i<m;i++)        {            if(!vis[j][i])            {                a=j;                b=i;                memset(vis,0,sizeof(vis));                dfs(j,i,0);            }        }    }    if(s) printf("Yes\n");    else printf("No\n");    return 0;}