codeforces 510.B Fox And Two Dots （DFS好题）

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
4000
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：有多种颜色，问你颜色相同的点能否组成一个环（只能上下左右的走）

本题用DFS，不同的是每次递归，要记录上一个走过点点，保证下次不能再走在个点，这样一直走，如果能走到标记过的点，那么就说明存在回路。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int px[4]={0,0,1,-1};//方向
int py[4]={-1,1,0,0};
int vis[55][55];//标记
char map[55][55];
int n,m,ans;
void DFS(int x,int y,int cx,int cy)//cx，cy记录上次走过的点
{
char c;
c=map[x][y];
for(int i=0;i<4;i++)
{
int nx=x+px[i];
int ny=y+py[i];
if(map[nx][ny]==c&&nx>=0&&nx<n&&ny>=0&&ny<m)
{
if(nx==cx&&ny==cy)  //不走上次走过的点
{
continue;
}
if(vis[nx][ny])  //如果它被标记，就说明存在回路
{
ans=1;
return ;
}
vis[nx][ny]=1;
DFS(nx,ny,x,y);
}
}
return ;
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
ans=0;
for(int i=0;i<n;i++)
scanf("%s",map[i]);
memset(vis,0,sizeof(vis));
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
if(!vis[i][j])
{
vis[i][j]=1;
DFS(i,j,i,j);
}
if(ans)
break;
}
if(ans)
break;
}
if(ans)
printf("Yes\n");
else printf("No\n");
}
return 0;
}