Fox And Two Dots

B - Fox And Two Dots

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status

Description

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, ..., d_k a cycle if and only if it meets the following condition:

1. These k dots are different: if i ≠ j then d_i is different from d_j.
2. k is at least 4.
3. All dots belong to the same color.
4. For all 1 ≤ i ≤ k - 1: d_i and d_i + 1 are adjacent. Also, d_k 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 Input

Input

3 4AAAAABCAAAAA

Output

Yes

Input

3 4AAAAABCAAADA

Output

No

Input

4 4YYYRBYBYBBBYBBBY

Output

Yes

Input

7 6AAAAABABBBABABAAABABABBBABAAABABBBABAAAAAB

Output

Yes

Input

2 13ABCDEFGHIJKLMNOPQRSTUVWXYZ

Output

No

Hint

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).

欧拉回路，用bfs搞了一通，发现好乱，学习大神DFS，听房神说还有很简单的方法，明天继续看一下。

 1 #include <iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<queue>
 6 using namespace std;
 7 char maze[55][55];
 8 int vis[55][55];
 9 int n,m;
10 int flag = 0;
11 int to[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};
12 bool check(int x,int y){
13     if(x<0||x>=n||y<0||y>=m) return false;
14     //if(vis[x][y]) return false;
15     return true;
16 }
17 void dfs(int x,int y,int prex,int prey){
18     if(!check(x,y)) return;
19     vis[x][y] = 1;
20     int postx,posty;
21     for(int i = 0; i<4; i++){
22         postx = x + to[i][0];
23         posty = y + to[i][1];
24         if(check(postx,posty)&&maze[postx][posty] == maze[x][y]&&(postx!=prex||posty!=prey)){
25             if(vis[postx][posty]){
26                 flag = 1;
27                 return;
28             }
29             dfs(postx,posty,x,y);
30         }
31     }
32 }
33 void input(){
34
35     scanf("%d%d",&n,&m);
36     for(int i = 0; i<n; i++) scanf("%s",maze[i]);
37     for(int i = 0; i<n; i++){
38         for(int j = 0; j<m; j++){
39             if(!vis[i][j]){
40                 dfs(i,j,-1,-1);
41             }
42         }
43     }
44     if(flag) printf("Yes\n");
45     else printf("No\n");
46 }
47 int main()
48 {
49     input();
50     return 0;
51 }

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