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.
Examples
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).
题意大概是，判断相同的字母是否能连成环。用dfs扫一遍整个图即可。
AC代码如下：
#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<string>
#include<map>
#include<algorithm>
#include<stack>
#include<queue>
using namespace std;
typedef long long ll;
const int maxn = 2e3+10;
char str[maxn][maxn];
int visb[maxn][maxn];
int to[4][2]={0,1,0,-1,1,0,-1,0};
int flag=0;
int n,m;
struct node
{
int x,y;
};
void bfs(int stx,int sty,int fx,int fy)
{
if(visb[stx][sty])
{
flag=1;
return;
}
visb[stx][sty]=1;
int nx,ny;
for(int i=0;i<4;i++)
{
nx=stx+to[i][0];
ny=sty+to[i][1];
if(nx==fx&&ny==fy)
continue;
if(str[nx][ny]==str[fx][fy]&&nx<n&&ny<m&&nx>=0&&ny>=0)
bfs(nx,ny,stx,sty);
}
}
int main()
{
scanf("%d %d",&n,&m);
flag=0;
for(int i=0;i<n;i++)
scanf("%s",str[i]);
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
if(!visb[i][j])
bfs(i,j,i,j);
if(flag)
break;
}
if(flag)
break;
}
if(!flag) printf("No\n");
else printf("Yes\n");
return 0;
}