D - Fox And Two Dots (codeforces 510B)

 

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:

1. These k dots are different: if i ≠ j then di is different from dj.
2. k is at least 4.
3. All dots belong to the same color.
4. 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).

 

 

题意：在一个棋盘里有不同种类的点，用大写的英文字母表示，问你在这个棋盘里是否有同种颜色的点构成的环，构成环的要求：每个点都要相邻并且第一个点要与最后一个点相邻；

思路：我用的dfs写的。。。总体思路就是从一个点出发，当搜索到尽头时就判断是否符合形成环的条件，如果成立就直接返回，不用继续搜索，如不满足条件，那么就将ans清0，否则函数回溯以后的步数将会和之前搜索的步数进行累加，导致结果错误。。。（因为这个错了3遍QAQ);

下面附上我的代码：

 

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
using namespace std;
int sx,sy,ex,ey,ny,nx,n,m,ans,ams,flag=0;
int mx[4]={0,0,1,-1};
int my[4]={1,-1,0,0};
int vis[205][205];
char mapp[205][205];
bool check(int x,int y)//判断是否是合法的点 
{
	if(x<0||y<0||x>=n||y>=m||vis[x][y])
		return false;
	return true;
}
void dfs(int x,int y)
{
	ans++;//记录步数 
	vis[x][y]=1;
	//printf("======%d %d %d %d======\n",sx,sy,ans,ams);
	for(int i=0;i<4;i++)
	{
		nx=x+mx[i];
		ny=y+my[i];
		if(check(nx,ny))
		{
			if(mapp[nx][ny]==mapp[x][y])//如果满足条件就继续搜索 
				dfs(nx,ny);
		}
	}
	if(ans>=4&&((abs(ex-x)==1&&abs(y-ey)==0)||(abs(ex-x)==0&&abs(y-ey)==1)))//判断搜到尽头的那个点是否和起点相邻，步数是否大于4 
	{
		flag=1;
		return;
	}
	else ans=0;//如果不满足条件即回溯到起点重新搜索，这时步数就要清零 
}
int main()
{
	while(~scanf("%d %d",&n,&m)&&(n||m))
	{
		flag=0;
		for(int i=0;i<n;i++)
			scanf("%s",mapp[i]);
		for(int i=0;i<n;i++)
			for(int j=0;j<m;j++)
			{	
				ans=0;
				memset(vis,0,sizeof(vis));//每搜一个点，判断完后都要清除标记，以便后面的点的搜索 
				if(!vis[i][j])
				{
					ex=i;
					ey=j;
					dfs(i,j);
					if(flag) break;
				}
				if(flag) break;
			}	
		if(flag) puts("Yes");
		else puts("No");
	}
	return 0;
}