CodeForces 510B Fox And Two Dots(深搜DFS)

题目：

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).

思路：

跟走迷宫差不多，注意标记上一步的，最后要走回来

代码:

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <string>
#include <iostream>
#include <stack>
#include <queue>
#include <vector>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
#define N 600+20
#define M 200010
#define MOD 1000000000+7
#define inf 0x3f3f3f3f
using namespace std;
char map[200][200];
int m,n,flag,t;
int go[4][2]= {{0,1},{-1,0},{0,-1},{1,0}},vis[200][200];
void dfs(int x,int y,int fx,int fy)
{
int i,j,xx,yy;
for(i=0; i<4; i++)
{
xx=x+go[i][0];
yy=y+go[i][1];
if(xx>=0&&xx<n&&yy>=0 && yy<m && map[xx][yy]==map[x][y])
{
if(xx==fx && yy==fy)
{
continue;
}
if(vis[xx][yy]==1)
{
flag=1;
break;
}
vis[xx][yy]=1;
dfs(xx,yy,x,y);
}
if(flag==1)
break;
}
return;
}
int main()
{
int i,j,u,v;
while(scanf("%d%d",&n,&m)!=EOF)
{
for(i=0; i<n; i++)
scanf("%s",map[i]);
mem(vis,0);
for(i=0; i<n; i++)
{
for(j=0; j<m; j++)
{
if(vis[i][j]==1)
continue;
flag=0;
vis[i][j]=1;
dfs(i,j,-1,-1);
if(flag==1)
break;
}
if(flag==1)
break;
}
if(flag)
printf("Yes\n");
else
printf("No\n");
}
return 0;
}