codeforce-B. Fox And Two Dots

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).

一题我用的是dfs 其实仔细想想，你会发现从一个点跑，你可以想象一下其实dfs就像一只会分身 的老鼠，到达一个点只要四周可以跑，就会分身，一直跑下去，只要存在一个回路，它就会发现已经走过 然后就会汇报有回路，如果跑到最后都没有回路，说明从这个点和所有跑过的路都不会有回路，因为只要 存在回路，它就会发现其他老鼠跑过的路 值得注意的时不要走回路   <span style="font-family:SimSun;">#include<cstdio> #include<cstring> char s[100][100]; int vis[100][100];//用于标记 int n,m;//n行m列 int flag; int dx[4]={1,-1,0,0}; int dy[4]={0,0,1,-1}; int judge(int a,int b) { if(a>=0&&a<n&&b>=0&&b<m) return 1; return 0; } void dfs(int x,int y,char o,int xx,int yy)//记录上一步走的路 { int i; for(i=0;i<4;++i) { int ex=x+dx[i]; int ey=y+dy[i]; if(judge(ex,ey)&&s[ex][ey]==o) { if(ex==xx&&ey==yy) continue; if(vis[ex][ey]) { flag=1; return ; } vis[ex][ey]=1; dfs(ex,ey,o,x,y); } } } int main() { scanf("%d%d",&n,&m); int i,j; for(i=0;i<n;++i) scanf("%s",s[i]);//这样就很容易形成地图 memset(vis,0,sizeof(vis)); for(i=0;i<n;++i) for(j=0;j<m;++j) { if(!vis[i][j]) { flag=0; vis[i][j]=1; dfs(i,j,s[i][j],-1,-1); if(flag) { printf("Yes\n"); return 0; } } } printf("No\n"); return 0; } </span>