CF B. Fox And Two Dots

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:

1. These k dots are different: if i ≠ j then di is different from dj.
2. k is at least 4.
3. All dots belong to the same color.
4. 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).

 题意：判断图中是否存在长度大于4的回路(由同一种字母组成的)。

思路：DFS，注意每次不往回搜索，直到搜索结束或者搜到有标记的位置停止。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<queue>
#include<algorithm>
#include<map>
#include<iomanip>
#include<climits>
#include<string.h>
#include<numeric>
#include<cmath>
#include<stdlib.h>
#include<vector>
#include<stack>
#include<set>
#define FOR(x, b, e)  for(int x=b;x<=(e);x++)
#define REP(x, n)     for(int x=0;x<(n);x++)
#define INF 1e7
#define MAXN 100010
#define maxn 1000010
#define Mod 1000007
#define N 1010
using namespace std;
typedef long long LL;


char G[55][55];
int vis[55][55];
int dx[] = {-1, 0, 0, 1}, dy[] = {0, 1, -1, 0};
int n, m;
bool ok;

void dfs(int x,int y, int d) 
{
    vis[x][y] = 1;
    REP(i, 4) {
        if (i == 3 - d) continue;
        int nx = x + dx[i];
        int ny = y + dy[i];
        if (nx == 0 || ny == 0 || ny == m+1 || nx == n+1) continue;
        if (G[x][y] == G[nx][ny]) {
            if (vis[nx][ny]) {
                ok = true;
                return;
            }
            else dfs(nx, ny, i);
        }
    }
}

int main()
{
    cin >> n >> m;
    FOR(i, 1, n)
        cin >> G[i] + 1;
    FOR(i, 1, n)
        FOR(j, 1, m) {
            if (!vis[i][j]) 
                dfs(i, j, 4);
        }
    puts(ok ? "Yes" : "No");
    return 0;
}