Codeforces Round #290 (Div. 2) B. 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.

Sample test(s)

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).

同一个字母是一种颜色，问从某一个字母，遍历与它相同的字母，问能否回到这个点，即是否有环。。

用DFS从某个字母开始遍历，记录下每个点的步数，如果一个点访问过了，回溯到某点时，相邻一点已访问过&&这一点的步数大于回溯的点的步数，则说明之前已经到达过，现在又能到达，则说明有环

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stdlib.h>
#include <algorithm>
#include <cmath>
#define MAX 0x3f3f3f3f
using namespace std;
char s[60][60];
int b[60][60];
int step[60][60];
int yy[]= {0,0,-1,1};
int xx[]= {1,-1,0,0};
int n,m,flag;
void dfs(int x,int y,char r)
{
    b[x][y]=1;
    for(int i=0; i<4; i++)
    {
        int dx=x+xx[i];
        int dy=y+yy[i];
        if(dx>=0 && dx<n &&dy>=0 && dy<m && r==s[dx][dy])
        {
            if(b[dx][dy]==0)
            {
                b[dx][dy]=1;
                step[dx][dy]=step[x][y]+1;
                dfs(dx,dy,r);
            }

            else if(b[dx][dy]&&(step[dx][dy]-step[x][y]>=1))
            {
                flag=1;
                return ;
            }
        }
    }
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=0; i<n; i++)
            scanf("%s",s[i]);
        memset(b,0,sizeof(b));
        memset(step,0,sizeof(step));
        flag=0;
        for(int i=0; i<n; i++)
        {
            for(int j=0; j<m; j++)
            {
                if(!b[i][j])
                {
                    b[i][j]=1;
                    step[i][j]=1;
                    dfs(i,j,s[i][j]);
                }
                if(flag)
                    break;
            }
            if(flag) break;
        }
        if(flag)
            printf("Yes\n");
        else printf("No\n");
    }
    return 0;
}