Fox And Two Dots
题目来源于2017HPUACM暑期培训：https://vjudge.net/contest/174968#problem/D
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.
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).
也是dfs的运用，但是注意这里面dfs不能回头，加变量判断它不走重复路线
#includeif(xx==px&&yy==py)//px，py判断是否回头
{
continue;
}
else
{
if(vis[xx][yy]&&s[xx][yy]==ss)
{
k=1;
return ;
}
}
vis[xx][yy]=1;
dfs(xx,yy,x,y,s[xx][yy]);
}
}
}
int main()
{
k=0;
memset(vis,0,sizeof(vis));
scanf("%d%d",&n,&m);
for(int i=0;iscanf("%s",s[i]);
for(int i=0;ifor(int j=0;jif(!vis[i][j])
{
vis[i][j]=1;
dfs(i,j,-1,-1,s[i][j]);
if(k==1)
break;
}
}
if(k==1)
break;
}
if(k==1)
printf("Yes\n");
else
printf("No\n");
return 0;
}