DFS CodeForces 510B （Fox And Two Dots）

   Fox And Two Dots

  

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 题意：给出字母矩阵，求相同字符能否成环。 题解：DFS，但是要注意回溯条件，也就是父节点不能再走。回溯后找到了已经标记过得点，说明成环。 #include<cstdio> #include<cstring> using namespace std; char map[55][55]; int vis[55][55]; int dx[4]={0,1,0,-1}; int dy[4]={1,0,-1,0}; int n,m; int i,j; int flag; void dfs(int a,int b,int aa,int bb){//aa,bb为父节点 int k; char c=map[a][b]; for (k=0;k<4;k++){ int x=a+dx[k]; int y=b+dy[k]; if (map[x][y]==c&&x>=0 &&y>=0 &&x<n&&y<m){ if (x==aa && y==bb) continue; if (vis[x][y]){ flag =1; return ; } vis[x][y]=1; dfs(x,y,a,b); } } } int main(){ scanf ("%d %d",&n,&m); flag=0; memset (vis,0,sizeof (vis)); for (i=0;i<n;i++){ scanf ("%s",map[i]); } for (i=0;i<n;i++){ for (j=0;j<m;j++){ if (!vis[i][j]) dfs(i,j,-1,-1); if (flag) break; } } if (flag) printf ("Yes\n"); else printf ("No\n"); return 0; }