B. Fox And Two Dots Codeforces Round #290 (Div. 2)

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:

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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.

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).

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <string> #include <map> #include <stack> #include <vector> #include <set> #include <queue> #define maxn 55 using namespace std; struct Node { int x,y,step,d; }; int dir[4][2]={-1,0,0,1,1,0,0,-1}; char mp[maxn][maxn]; int step[maxn][maxn]; bool vis[maxn][maxn]; int n,m; bool isok(int x,int y) { if (x>=0&&x<n&&y>=0&&y<m) return true; return false; } bool bfs(int x,int y) { Node st,now; queue<Node>Q; while (!Q.empty()) Q.pop(); st.x=x;st.y=y;st.step=0;st.d=-1; Q.push(st); step[x][y]=0; vis[x][y]=true; while (!Q.empty()) { st=Q.front(); Q.pop(); for (int i=0;i<4;i++) { if (st.d==(i+2)%4) continue; //不能走回头路 now.x=st.x+dir[i][0]; now.y=st.y+dir[i][1]; now.step=st.step+1; now.d=i; //记录当前点走到下一个点是从哪个方向过去的 if (isok(now.x,now.y)&&mp[st.x][st.y]==mp[now.x][now.y]) { if (vis[now.x][now.y]&&(now.step+step[now.x][now.y]+1)>=4) //碰头了并且长度之和大于4 return true; if (!vis[now.x][now.y]) //没有被访问过就入队列 { vis[now.x][now.y]=true; step[now.x][now.y]=now.step; Q.push(now); } } } } return false; } bool solve() { for (int i=0;i<n;i++) for (int j=0;j<m;j++) if (!vis[i][j]&&bfs(i,j)) return true; return false; } int main() { while (~scanf("%d%d",&n,&m)) { for (int i=0;i<n;i++) scanf("%s",mp[i]); memset(vis,false,sizeof(vis)); memset(step,0,sizeof(step)); if (solve()) printf("Yes\n"); else printf("No\n"); } return 0; }