Codeforces 510B. Fox And Two Dots

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:

These k dots are different: if i ≠ j then d_i is different from d_j. k is at least 4. All dots belong to the same color. 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 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

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).

这道题可以用dfs来做，从起点出发开始搜索，当再次搜索到起点的时候并且所用步骤大于等于4的时候就返回；

#include <iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<algorithm> #define INF 0xfffffff using namespace std; char s[100][100]; int b[100][100],flag,n,m,step,p,q; void dfs(int x,int y) { int next[4][2]= {{0,1},{1,0},{0,-1},{-1,0}}; int k,tx,ty; for(k=0; k<=3; k++) { tx=x+next[k][0]; ty=y+next[k][1]; if(tx>=0&&ty>=0&&tx<n&&ty<m&&s[tx][ty]==s[x][y]) { if(b[tx][ty]==0) { step++;//step来记录走过的步骤是多少 b[tx][ty]=1;//将走过的点标记一下，注意走过的路如果搜索不成功的话就不需要再走，所以回溯的不需要再置为零； dfs(tx,ty); step--;//回溯的时候记得要把走过的步骤再一步步减去； } if(tx==p&&ty==q&&step>=4)//当再次碰见起点并且步骤大于4的时候，标记一下，然后结束搜索； { flag=1; return; } } } return; } int main() { int i,j; while(~scanf("%d%d",&n,&m)) { for(i=0; i<n; i++) scanf("%s",s[i]); for(i=0; i<n; i++) { for(j=0; j<m; j++) { memset(b,0,sizeof(b));//每次都要初始化数组 b[i][j]=1; flag=0; step=1; p=i;//记录起点 q=j; dfs(i,j); if(flag) break; } if(flag) break; } if(flag) printf("Yes\n"); else printf("No\n"); } return 0; }