B. Fox And Two Dots

题目连接：

http://codeforces.com/contest/510/problem/B

Description

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 Input

3 4

AAAA

ABCA

AAAA

Sample Output

Yes

Hint

题意

给你一个n*m的网格

然后问你是否有只含有一种元素的环

题解：

dfs就好了

dfs的时候，记录一下fa，然后一直跑下去，跑到曾经vis过的地方，就说明遇到了环

代码

#include<bits/stdc++.h>

using namespace std;

char mp[55][55];

int vis[55][55];

int flag = 0;

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

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

int n,m;

void dfs(int x,int y,char c,int fax,int fay)

{

    vis[x][y]=1;

    if(flag)return;

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

    {

        int xx = x+dx[i];

        int yy = y+dy[i];

        if(xx==fax&&yy==fay)continue;

        if(xx<0||xx>=n)continue;

        if(yy<0||yy>=m)continue;

        if(mp[xx][yy]!=c)continue;

        if(vis[xx][yy]){

            flag=1;

            return;

        }

        dfs(xx,yy,c,x,y);

    }

}

int main()

{

    scanf("%d%d",&n,&m);

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

        scanf("%s",&mp[i]);

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

        for(int j=0;j<m;j++)

        {

            if(!vis[i][j])

                dfs(i,j,mp[i][j],i,j);

        }

    if(flag)printf("Yes");

    else printf("No\n");

}