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.

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 从一个点出发，向它的四个方向走，如果从上一点走过来不能再走上一个点，看到最后是否能回到原点 #include<stdio.h> #include<string.h> int visit[55][55],n,m; char map[55][55]; int yes; int move[4][2]={0,1,0,-1,1,0,-1,0}; void dfs(int x,int y,int a,int b) { if(yes) return ; for(int i=0;i<4;i++) { int ax=x+move[i][0]; int ay=y+move[i][1]; if(ax>=0&&ax<n&&ay>=0&&ay<m&&(ax!=a||ay!=b)&&map[ax][ay]==map[x][y]) { if(visit[ax][ay]) { yes=1; return; } visit[ax][ay]=1; dfs(ax,ay,x,y); } } } int main() { int i,j; while(~scanf("%d%d",&n,&m)) { for(i=0;i<n;i++) scanf("%s",map[i]); yes=0; for(i=0;i<n;i++) for(j=0;j<m;j++) { memset(visit,0,sizeof(visit)); visit[i][j]=1; dfs(i,j,i,j); if(yes) break; } if(yes) printf("Yes\n"); else printf("No\n"); } return 0; }