CodeForces 510B Fox And Two Dots （DFS）

B. Fox And Two Dots

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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).
只能说自己对DFS不够理解吧，才致使不能灵活运用
题意：判断是否有字母成环。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int dx[]={0,0,1,-1};
int dy[]={-1,1,0,0};
char map[55][55];
int vis[55][55];
int n,m;
bool dfs(int x,int y,int sx,int sy,char c)
{
vis[x][y]=1;
for(int i=0;i<4;i++)
{
int tx=x+dx[i];
int ty=y+dy[i];
if(tx==sx&&ty==sy)
continue;
if(tx>=0&&ty>=0&&tx<n&&ty<m&&map[tx][ty]==c)
{
if(vis[tx][ty]==1)
return 1;
if(dfs(tx,ty,x,y,c))
return 1;
}

}
return 0;
}
int main()
{
int i,j;
while(scanf("%d%d",&n,&m)!=EOF)
{
for(i=0;i<n;i++)
scanf("%s",map[i]);
for(i=0;i<n;i++)
for(j=0;j<m;j++)
{
if(!vis[i][j])
{
if(dfs(i,j,-1,-1,map[i][j]))
{
printf("Yes\n");
return 0;
}
}

}
printf("No\n");
}
return 0;
}