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).
主要是用dfs判断有没有回到出发的点，并保证不走回头路（所以dfs函数里定义四个变量）
#include<cstdio>
#include<cstring>
int flag[][];
char str[][];
int n,m,k;
int dx[]={-,,,};
int dy[]={,,-,};
void dfs(int x,int y,int cx,int cy)
{
if(flag[x][y] == )
{
k=;
return ;
}
flag[x][y]=;
int i,nx,ny;
for(i = ; i < ; i++)
{ nx=x+dx[i];
ny=y+dy[i];
if(str[nx][ny] == str[x][y] && (nx != cx || ny != cy))
{ dfs(nx,ny,x,y); }
} }
int main()
{
while(scanf("%d %d",&n,&m)!=EOF)
{
k=;
int i,j;
memset(flag,,sizeof(flag));
for(i = ; i <= n ; i++)
{
scanf("%s",str[i]+);
}
for(i = ; i <= n ; i++)
{
for(j = ; j <= m ; j++)
{
if(flag[i][j] == ) continue; dfs(i,j,i,j); if(k == ) break;
}
if(k == )
break;
}
if(k == )
printf("Yes\n");
else
printf("No\n");
}
}