Codeforces 510B：Fox And Two Dots（DFS变形+技巧）

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)

题目大意：给你一张字母图，让你判断其中存在某种字母成环。

解题思路：DFS，既然让成环，那么最终又得回到起点，但是这种是不行的AA（从第一个A搜到第二个A，而第二个A又搜到了第一个A，显然不符合题意）。那么这里我们多加两个参数，来记载当前节点的父节点（即它从哪来的），当你扫下一个方向时，扫到父节点，那么就跳过，扫下一个方向，如果你找到了你曾经走过的点，那么一定成环了。总结下，就是不走回头路的找到了同胞，那么就成环了。

代码如下：

#include <cstdio>
#include <cstring>
char map[55][55];
int visit[55][55];
int dirn[4]={1,-1,0,0};
int dirm[4]={0,0,1,-1};
int n,m;
int flag;//标记是否成环
int fx,fy;
void dfs(int hang,int lie,int fn,int fm,char word)//fn、fm代表当前点的父节点
{
if(flag==1)//找到环了，那么之前的每层递归都快速结束 节约时间
return ;
for(int i=0;i<4;i++)
{
int nx=hang+dirn[i];
int ny=lie+dirm[i];
if(nx>=0&&ny>=0&&nx<n&&ny<m&&map[nx][ny]==word)//不用判断之前是否走过，因为排除了父节点再碰到之前走过的点的话，那么一定成环了
{
if(nx==fn&&ny==fm)//跳过父节点
{
continue;
}
if(visit[nx][ny]==1)//成环
{
flag=1;
return ;
}
visit[nx][ny]=1;//标记下
dfs(nx,ny,hang,lie,word);//让当前点做下一个点的父节点
}
}
}
int main()
{
scanf("%d%d",&n,&m);
for(int i=0;i<n;i++)
{
scanf("%s",map[i]);
}
memset(visit,0,sizeof(visit));
flag=0;//初始化标记
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
fx=-5;//初始化当前点的父节点 ，-5肯定扫不到
fy=-5;
if(visit[i][j]==0)//之前标记的就不扫了
{
visit[i][j]=1;
dfs(i,j,fx,fy,map[i][j]);
if(flag==1)
{
break;
}
}
}
if(flag==1)
{
break;
}
}
if(flag==1)
{
printf("Yes\n");
}
else
{
printf("No\n");
}
return 0;
}