字符矩阵中是否存在环

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 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).

#include<cstdio> #include<cstring> #include<algorithm> using namespace std; int n,m,cot; char c[110][110]; int vis[110][110]; int dx[4]={1,-1,0,0}; int dy[4]={0,0,1,-1}; bool dfs(char s,int sx,int sy,int x,int y) { if(sx==x&&sy==y&&vis[sx][sy]) return cot>=4?true:false;//围成环，至少4个 vis[x][y]=1; for(int i=0;i<4;i++) { int tx=x+dx[i]; int ty=y+dy[i]; if(tx>=0&&tx<n&&ty>=0&&ty<m&&c[tx][ty]==s&&vis[tx][ty]==0||(tx==sx&&ty==sy)) { cot++;//记录相同字母个数 if(dfs(s,sx,sy,tx,ty))//内循环,深搜 return true; cot--;//恢复原值，方便记录下个字母 } } return false; } bool solve() { for(int i=0;i<n;i++) { for(int j=0;j<m;j++) { memset(vis,0,sizeof(vis)); cot=0; if(dfs(c[i][j],i,j,i,j)) return true; } } return false; } int main() { while(~scanf("%d%d",&n,&m)) { for(int i=0;i<n;i++) scanf("%s",&c[i]); if(solve()) printf("Yes\n"); else printf("No\n"); } return 0; }